Math 107 Practice Final Test, Solution

1. f(x)=3 x 2 −2x+5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaaiodacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaadIhacqGHRaWkcaaI1aaaaa@4129@ .

So, f(2x+3)=3 (2x+3) 2 −2(2x+3)+5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaGaeyypa0JaaG4maiaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaGaey4kaSIaaGynaaaa@4AEC@

= 3(4 x 2 +12x+9)−2(2x+3)+5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4maiaacIcacaaI0aGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdacaaIYaGaamiEaiabgUcaRiaaiMdacaGGPaGaeyOeI0IaaGOmaiaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaGaey4kaSIaaGynaaaa@47A8@

= 12 x 2 +36x+27−4x−6+5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maiaaiAdacaWG4bGaey4kaSIaaGOmaiaaiEdacqGHsislcaaI0aGaamiEaiabgkHiTiaaiAdacqGHRaWkcaaI1aaaaa@4506@

= 12 x 2 +32x+26 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maiaaikdacaWG4bGaey4kaSIaaGOmaiaaiAdaaaa@3F0B@

 

2. f(x)= 4 x 2 −4x−3 = (2x−3)(2x+1) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maakaaabaGaaGinaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabgkHiTiaaiodaaSqabaGccqGH9aqpdaGcaaqaaiaacIcacaaIYaGaamiEaiabgkHiTiaaiodacaGGPaGaaiikaiaaikdacaWG4bGaey4kaSIaaGymaiaacMcaaSqabaaaaa@4BE6@

For f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  to be defined, we should be able to determine the square root. The square root is defined for numbers that are positive or 0. So, we solve

(2x−3)(2x+1)≥0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaWG4bGaeyOeI0IaaG4maiaacMcacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIXaGaaiykaiabgwMiZkaaicdaaaa@41D9@

First, we note that (2x−3)(2x+1)=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaWG4bGaeyOeI0IaaG4maiaacMcacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIXaGaaiykaiabg2da9iaaicdaaaa@4119@  if x= 3 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaaaaa@397A@  or x= −1 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9maalaaabaGaeyOeI0IaaGymaaqaaiaaikdaaaaaaa@3A65@ . These two numbers divide the real line into 3 intervals, in which the sign of (2x−3)(2x+1) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaWG4bGaeyOeI0IaaG4maiaacMcacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIXaGaaiykaaaa@3F59@  is same. By picking points and checking the sign, we note that (2x−3)(2x+1)≥0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaWG4bGaeyOeI0IaaG4maiaacMcacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIXaGaaiykaiabgwMiZkaaicdaaaa@41D9@  on the interval ( −∞, −1 2 ]∪[ 3 2 ,∞ ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaKamaeaacqGHsislcqGHEisPcaGGSaWaaSaaaeaacqGHsislcaaIXaaabaGaaGOmaaaaaiaawIcacaGLDbaacqWIQisvdaqcsaqaamaalaaabaGaaG4maaqaaiaaikdaaaGaaiilaiabg6HiLcGaay5waiaawMcaaaaa@4411@ .

So, the domain = ( −∞, −1 2 ]∪[ 3 2 ,∞ ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaKamaeaacqGHsislcqGHEisPcaGGSaWaaSaaaeaacqGHsislcaaIXaaabaGaaGOmaaaaaiaawIcacaGLDbaacqWIQisvdaqcsaqaamaalaaabaGaaG4maaqaaiaaikdaaaGaaiilaiabg6HiLcGaay5waiaawMcaaaaa@4411@ .

 

3. f(x)= x 2 +5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI1aaaaa@3DC6@ .

So, lim⁡ h→0 f(2x+h)−f(2x) h MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIgacqGHsgIRcaaIWaaabeaakmaalaaabaGaamOzaiaacIcacaaIYaGaamiEaiabgUcaRiaadIgacaGGPaGaeyOeI0IaamOzaiaacIcacaaIYaGaamiEaiaacMcaaeaacaWGObaaaaaa@4848@  = lim⁡ h→0 [ (2x+h) 2 +5 ]−[ (2x) 2 +5 ] h MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIgacqGHsgIRcaaIWaaabeaakmaalaaabaWaamWaaeaacaGGOaGaaGOmaiaadIhacqGHRaWkcaWGObGaaiykamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiwdaaiaawUfacaGLDbaacqGHsisldaWadaqaaiaacIcacaaIYaGaamiEaiaacMcadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI1aaacaGLBbGaayzxaaaabaGaamiAaaaaaaa@4F7E@

= lim⁡ h→0 4 x 2 +4xh+ h 2 +5−4 x 2 −5 h MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIgacqGHsgIRcaaIWaaabeaakmaalaaabaGaaGinaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI0aGaamiEaiaadIgacqGHRaWkcaWGObWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGynaiabgkHiTiaaisdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGynaaqaaiaadIgaaaaaaa@4D74@

= lim⁡ h→0 4xh+ h 2 h MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIgacqGHsgIRcaaIWaaabeaakmaalaaabaGaaGinaiaadIhacaWGObGaey4kaSIaamiAamaaCaaaleqabaGaaGOmaaaaaOqaaiaadIgaaaaaaa@42FC@  = lim⁡ h→0 (4x+h)=4x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIgacqGHsgIRcaaIWaaabeaakiaacIcacaaI0aGaamiEaiabgUcaRiaadIgacaGGPaGaeyypa0JaaGinaiaadIhaaaa@4439@ .

 

4. f(x)=2x (3x−11) 2 (2−5x)(x+5) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaaikdacaWG4bGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaOGaaiikaiaaikdacqGHsislcaaI1aGaamiEaiaacMcacaGGOaGaamiEaiabgUcaRiaaiwdacaGGPaaaaa@4B0C@ .

f(x)=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaaicdaaaa@3AEF@  if x=0, 11 3 , 2 5 ,−5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9iaaicdacaGGSaWaaSaaaeaacaaIXaGaaGymaaqaaiaaiodaaaGaaiilamaalaaabaGaaGOmaaqaaiaaiwdaaaGaaiilaiabgkHiTiaaiwdaaaa@4035@ . These 4 points divide the real line into 5 intervals in which the sign of f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  is same. We pick points in each interval to determine the signs. Actually, once we know the sign in one interval, we can determine the sign in other intervals by considering the exponents of each linear factor. Below is the sign graph of f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@

5. f(x)=−2 x 2 +7x+11 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iabgkHiTiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4naiaadIhacqGHRaWkcaaIXaGaaGymaaaa@42C6@ . So, f ′ (x)=−4x+7 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaeyOeI0IaaGinaiaadIhacqGHRaWkcaaI3aaaaa@3E8C@ . The slope of the tangent line to the graph of f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  at the point (-1,2) is same as f ′ (−1)=−4(−1)+7=4+7=11 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiabgkHiTiaaigdacaGGPaGaeyypa0JaeyOeI0IaaGinaiaacIcacqGHsislcaaIXaGaaiykaiabgUcaRiaaiEdacqGH9aqpcaaI0aGaey4kaSIaaG4naiabg2da9iaaigdacaaIXaaaaa@471E@ . So, the equation of the tangent line using the slop-point form y− y o =m(x− x 0 ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabgkHiTiaadMhadaWgaaWcbaGaam4BaaqabaGccqGH9aqpcaWGTbGaaiikaiaadIhacqGHsislcaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaaiykaaaa@4129@ , we have y−2=11(x+1) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabgkHiTiaaikdacqGH9aqpcaaIXaGaaGymaiaacIcacaWG4bGaey4kaSIaaGymaiaacMcaaaa@3F04@  or, y−2=11x+11 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabgkHiTiaaikdacqGH9aqpcaaIXaGaaGymaiaadIhacqGHRaWkcaaIXaGaaGymaaaa@3E66@  or y=11x+13 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2da9iaaigdacaaIXaGaamiEaiabgUcaRiaaigdacaaIZaaaaa@3CBF@ .

 

6.  f(x)= ( x 2 −x+1) 2 (2x+3) 3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgUcaRiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaOGaaiikaiaaikdacaWG4bGaey4kaSIaaG4maiaacMcadaahaaWcbeqaaiaaiodaaaaaaa@4793@ . We use the product rule to find the derivative.

f ′ (x)=[ 2( x 2 −x+1)(2x−1) ] (2x+3) 3 + ( x 2 −x+1) 2 [ 3 (2x+3) 2 (2) ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@612B@

= 2( x 2 −x+1)(2x−1) (2x+3) 3 +6 ( x 2 −x+1) 2 (2x+3) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgUcaRiaaigdacaGGPaGaaiikaiaaikdacaWG4bGaeyOeI0IaaGymaiaacMcacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIZaGaaiykamaaCaaaleqabaGaaG4maaaakiabgUcaRiaaiAdacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadIhacqGHRaWkcaaIXaGaaiykamaaCaaaleqabaGaaGOmaaaakiaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaWaaWbaaSqabeaacaaIYaaaaaaa@56D8@

= 2( x 2 −x+1) (2x+3) 2 [ (2x−1)(2x+3)+3( x 2 −x+1) ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgUcaRiaaigdacaGGPaGaaiikaiaaikdacaWG4bGaey4kaSIaaG4maiaacMcadaahaaWcbeqaaiaaikdaaaGcdaWadaqaaiaacIcacaaIYaGaamiEaiabgkHiTiaaigdacaGGPaGaaiikaiaaikdacaWG4bGaey4kaSIaaG4maiaacMcacqGHRaWkcaaIZaGaaiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWG4bGaey4kaSIaaGymaiaacMcaaiaawUfacaGLDbaaaaa@56EA@

= 2( x 2 −x+1) (2x+3) 2 [ (4 x 2 +4x−3)+(3 x 2 −3x+3) ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgUcaRiaaigdacaGGPaGaaiikaiaaikdacaWG4bGaey4kaSIaaG4maiaacMcadaahaaWcbeqaaiaaikdaaaGcdaWadaqaaiaacIcacaaI0aGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaisdacaWG4bGaeyOeI0IaaG4maiaacMcacqGHRaWkcaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIZaGaamiEaiabgUcaRiaaiodacaGGPaaacaGLBbGaayzxaaaaaa@568C@

= 2( x 2 −x+1) (2x+3) 2 (7 x 2 +x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgUcaRiaaigdacaGGPaGaaiikaiaaikdacaWG4bGaey4kaSIaaG4maiaacMcacaGGOaGaaG4naiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG4bGaaiykaaaa@4814@

= 2x( x 2 −x+1) (2x+3) 2 (7x+1) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaadIhacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadIhacqGHRaWkcaaIXaGaaiykaiaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaGaaiikaiaaiEdacaWG4bGaey4kaSIaaGymaiaacMcaaaa@47DC@

 

7. f(x)= (3x−1) 2 (2−5x) 4 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maalaaabaGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaakeaacaGGOaGaaGOmaiabgkHiTiaaiwdacaWG4bGaaiykamaaCaaaleqabaGaaGinaaaaaaaaaa@459C@ . We use the quotient rule to find the derivative.

f ′ (x)= [ 2(3x−1)(3) ] (2−5x) 4 − (3x−1) 2 [ 4 (2−5x) 3 (−5) ] [ (2−5x) 2 ] 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6411@

= 6(3x−1) (2−5x) 4 +20 (3x−1) 2 (2−5x) 3 (2−5x) 8 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaI2aGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaacMcacaGGOaGaaGOmaiabgkHiTiaaiwdacaWG4bGaaiykamaaCaaaleqabaGaaGinaaaakiabgUcaRiaaikdacaaIWaGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaGccaGGOaGaaGOmaiabgkHiTiaaiwdacaWG4bGaaiykamaaCaaaleqabaGaaG4maaaaaOqaaiaacIcacaaIYaGaeyOeI0IaaGynaiaadIhacaGGPaWaaWbaaSqabeaacaaI4aaaaaaaaaa@5491@

= 2(3x−1) (2−5x) 3 [ 3(2−5x)+10(3x−1) ] (2−5x) 8 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaacMcacaGGOaGaaGOmaiabgkHiTiaaiwdacaWG4bGaaiykamaaCaaaleqabaGaaG4maaaakmaadmaabaGaaG4maiaacIcacaaIYaGaeyOeI0IaaGynaiaadIhacaGGPaGaey4kaSIaaGymaiaaicdacaGGOaGaaG4maiaadIhacqGHsislcaaIXaGaaiykaaGaay5waiaaw2faaaqaaiaacIcacaaIYaGaeyOeI0IaaGynaiaadIhacaGGPaWaaWbaaSqabeaacaaI4aaaaaaaaaa@5553@

= 2(3x−1) (2−5x) 3 (6−15x+30x−10) (2−5x) 8 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaacMcacaGGOaGaaGOmaiabgkHiTiaaiwdacaWG4bGaaiykamaaCaaaleqabaGaaG4maaaakiaacIcacaaI2aGaeyOeI0IaaGymaiaaiwdacaWG4bGaey4kaSIaaG4maiaaicdacaWG4bGaeyOeI0IaaGymaiaaicdacaGGPaaabaGaaiikaiaaikdacqGHsislcaaI1aGaamiEaiaacMcadaahaaWcbeqaaiaaiIdaaaaaaaaa@5209@

= 2(3x−1)(15x−4) (2−5x) 5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaGaaiikaiaaiodacaWG4bGaeyOeI0IaaGymaiaacMcacaGGOaGaaGymaiaaiwdacaWG4bGaeyOeI0IaaGinaiaacMcaaeaacaGGOaGaaGOmaiabgkHiTiaaiwdacaWG4bGaaiykamaaCaaaleqabaGaaGynaaaaaaaaaa@469A@

 

8. (2 x 2 + y 2 ) 4 =2 y 2 + x 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyEamaaCaaaleqabaGaaGOmaaaakiaacMcadaahaaWcbeqaaiaaisdaaaGccqGH9aqpcaaIYaGaamyEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaaaa@4436@

Differentiating both sides with respect to x, we get

4 (2 x 2 + y 2 ) 3 (4x+2y y ′ )=4y y ′ +2x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeeG+aaaaaaivzKbWdbiaaisdacaGGOaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaG4maaaak8aacaGGOaaeeaaaOpa6dmuEH5gapiGaaGinaiaadIhapaGaey4kaSYdciaaikdacaWG5bGabmyEayaafaWdaiaacMcacqGH9aqpcaaI0aGaamyEaiqadMhagaqbaiabgUcaRiaaikdacaWG4baaaa@5187@

Or, 4 (2 x 2 + y 2 ) 3 (4x)+4 (2 x 2 + y 2 ) 3 (2y y ′ )=4y y ′ +2x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeeG+aaaaaaivzKbWdbiaaisdacaGGOaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaG4maaaak8aacaGGOaaeeaaaOpa6dmuEH5gapiGaaGinaiaadIhapaGaaiykaiabgUcaR8qacaaI0aGaaiikaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyEamaaCaaaleqabaGaaGOmaaaakiaacMcadaahaaWcbeqaaiaaiodaaaGcpaGaaiika8GacaaIYaGaamyEaiqadMhagaqba8aacaGGPaGaeyypa0JaaGinaiaadMhaceWG5bGbauaacqGHRaWkcaaIYaGaamiEaaaa@5B89@

Or, 16x (2 x 2 + y 2 ) 3 +8y y ′ (2 x 2 + y 2 ) 3 =4y y ′ +2x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaiAdacaWG4bGaaiikaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyEamaaCaaaleqabaGaaGOmaaaakiaacMcadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaI4aGaamyEaiqadMhagaqbaiaacIcacaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccaGGPaWaaWbaaSqabeaacaaIZaaaaOGaeyypa0JaaGinaiaadMhaceWG5bGbauaacqGHRaWkcaaIYaGaamiEaaaa@5211@

Or, 8y y ′ (2 x 2 + y 2 ) 3 −4y y ′ =2x−16x (2 x 2 + y 2 ) 3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGioaiaadMhaceWG5bGbauaacaGGOaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaisdacaWG5bGabmyEayaafaGaeyypa0JaaGOmaiaadIhacqGHsislcaaIXaGaaGOnaiaadIhacaGGOaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaG4maaaaaaa@521D@

Or, y ′ [ 8y (2 x 2 + y 2 ) 3 −4y ]=2x−16x (2 x 2 + y 2 ) 3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafaWaamWaaeaacaaI4aGaamyEaiaacIcacaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccaGGPaWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0IaaGinaiaadMhaaiaawUfacaGLDbaacqGH9aqpcaaIYaGaamiEaiabgkHiTiaaigdacaaI2aGaamiEaiaacIcacaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccaGGPaWaaWbaaSqabeaacaaIZaaaaaaa@5305@

Or, y ′ = 2x−16x (2 x 2 + y 2 ) 3 8y (2 x 2 + y 2 ) 3 −4y = x−8x (2 x 2 + y 2 ) 3 4y (2 x 2 + y 2 ) 3 −2y MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafaGaeyypa0ZaaSaaaeaacaaIYaGaamiEaiabgkHiTiaaigdacaaI2aGaamiEaiaacIcacaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccaGGPaWaaWbaaSqabeaacaaIZaaaaaGcbaGaaGioaiaadMhacaGGOaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaisdacaWG5baaaiabg2da9maalaaabaGaamiEaiabgkHiTiaaiIdacaWG4bGaaiikaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyEamaaCaaaleqabaGaaGOmaaaakiaacMcadaahaaWcbeqaaiaaiodaaaaakeaacaaI0aGaamyEaiaacIcacaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccaGGPaWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0IaaGOmaiaadMhaaaaaaa@69E7@

9. f(x)= e 2x e 3x +5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maalaaabaGaamyzamaaCaaaleqabaGaaGOmaiaadIhaaaaakeaacaWGLbWaaWbaaSqabeaacaaIZaGaamiEaaaakiabgUcaRiaaiwdaaaaaaa@419B@ . Using the quotient rule for derivative, we get

f ′ (x)= (2 e 2x )( e 3x +5)−( e 2x )(3 e 3x ) ( e 3x +5) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0ZaaSaaaeaacaGGOaGaaGOmaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaOGaaiykaiaacIcacaWGLbWaaWbaaSqabeaacaaIZaGaamiEaaaakiabgUcaRiaaiwdacaGGPaGaeyOeI0IaaiikaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaOGaaiykaiaacIcacaaIZaGaamyzamaaCaaaleqabaGaaG4maiaadIhaaaGccaGGPaaabaGaaiikaiaadwgadaahaaWcbeqaaiaaiodacaWG4baaaOGaey4kaSIaaGynaiaacMcadaahaaWcbeqaaiaaikdaaaaaaaaa@55E4@

= 2 e 5x +10 e 2x −3 e 5x ( e 3x +5) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaGaamyzamaaCaaaleqabaGaaGynaiaadIhaaaGccqGHRaWkcaaIXaGaaGimaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaOGaeyOeI0IaaG4maiaadwgadaahaaWcbeqaaiaaiodacaWG4baaaaGcbaGaaiikaiaadwgadaahaaWcbeqaaiaaiodacaWG4baaaOGaey4kaSIaaGynaiaacMcadaahaaWcbeqaaiaaikdaaaaaaaaa@4A0B@  = 10 e 2x − e 5x ( e 3x +5) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaGaaGimaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaOGaeyOeI0IaamyzamaaCaaaleqabaGaaG4maiaadIhaaaaakeaacaGGOaGaamyzamaaCaaaleqabaGaaG4maiaadIhaaaGccqGHRaWkcaaI1aGaaiykamaaCaaaleqabaGaaGOmaaaaaaaaaa@44D3@

 

10. y= x 2 ln⁡(4 x 2 −5x+7) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2da9iaadIhadaahaaWcbeqaaiaaikdaaaGcciGGSbGaaiOBaiaacIcacaaI0aGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiwdacaWG4bGaey4kaSIaaG4naiaacMcaaaa@4419@ . Using the product rule for derivative, we get

f ′ (x)=(2x)ln⁡(4 x 2 −5x+7)+( x 2 )( 8x−5 4 x 2 −5x+7 ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaaiikaiaaikdacaWG4bGaaiykaiGacYgacaGGUbGaaiikaiaaisdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGynaiaadIhacqGHRaWkcaaI3aGaaiykaiabgUcaRiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaabmaabaWaaSaaaeaacaaI4aGaamiEaiabgkHiTiaaiwdaaeaacaaI0aGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiwdacaWG4bGaey4kaSIaaG4naaaaaiaawIcacaGLPaaaaaa@57B3@

= 2xln⁡(4 x 2 −5x+7)+ x 2 (8x−5) 4 x 2 −5x+7 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaadIhaciGGSbGaaiOBaiaacIcacaaI0aGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiwdacaWG4bGaey4kaSIaaG4naiaacMcacqGHRaWkdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccaGGOaGaaGioaiaadIhacqGHsislcaaI1aGaaiykaaqaaiaaisdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGynaiaadIhacqGHRaWkcaaI3aaaaaaa@507E@

 

11. f ′ (x)= (x−1)(x+4) (x+3)(x−2) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0ZaaSaaaeaacaGGOaGaamiEaiabgkHiTiaaigdacaGGPaGaaiikaiaadIhacqGHRaWkcaaI0aGaaiykaaqaaiaacIcacaWG4bGaey4kaSIaaG4maiaacMcacaGGOaGaamiEaiabgkHiTiaaikdacaGGPaaaaaaa@4A39@ .

The critical points are where the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@  is zero or is undefined. Clearly, it is zero when the numerator (x−1)(x+4) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHsislcaaIXaGaaiykaiaacIcacaWG4bGaey4kaSIaaGinaiaacMcaaaa@3DE2@  is zero or at x = 1 and x = -4, and is undefined when the denominator (x+3)(x−2) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHRaWkcaaIZaGaaiykaiaacIcacaWG4bGaeyOeI0IaaGOmaiaacMcaaaa@3DE2@  is zero or at x = -3 and x = 2. So, the critical points are at x = 1, -4, -3 and 2.

These 4 critical points divide the real line into 5 intervals in which the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@  has the same sign. We can pick points in each interval and determine the signs or we find the sign in one interval and then by considering the exponent of each factor we find the signs in other intervals. Below is the sign graph of f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@

Note that where ever the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@  is positive, the function f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  is increasing, and where ever the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@  is negative, the function f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  is decreasing.

So, the function is increasing on (−∞,−4)∪(−3,1)∪(2,∞) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiabgkHiTiabg6HiLkaacYcacqGHsislcaaI0aGaaiykaiablQIivjaacIcacqGHsislcaaIZaGaaiilaiaaigdacaGGPaGaeSOkIuLaaiikaiaaikdacaGGSaGaeyOhIuQaaiykaaaa@4708@  and is decreasing on (−4,−3)∪(1,2) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiabgkHiTiaaisdacaGGSaGaeyOeI0IaaG4maiaacMcacqWIQisvcaGGOaGaaGymaiaacYcacaaIYaGaaiykaaaa@3FFE@

 

12. f ′ (x)=2 x 2 (x+2)(2x−5)(x−4) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccaGGOaGaamiEaiabgUcaRiaaikdacaGGPaGaaiikaiaaikdacaWG4bGaeyOeI0IaaGynaiaacMcacaGGOaGaamiEaiabgkHiTiaaisdacaGGPaaaaa@49A0@ . The critical points are at x = 0, -2, 4 and 5 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaI1aaabaGaaGOmaaaaaaa@3779@ . These four critical points divide the real line into 5 intervals in each of which the sign of the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@  is the same. We determine the signs by picking points in each interval and evaluating the derivative, or find the sign of the derivative in any one interval and then finding the sign in rest of the intervals by considering the exponents of each of the factors. Below is the sign graph of the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@ .

The function f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  is increasing when the derivative f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@  is positive, and is decreasing when the derivative is negative.

Using the first derivative test, which  says that if the function is decreasing on the left of a critical point and increasing on the right of that critical point, the function has a relative minimum, and that if the function is increasing on the left of the critical point and decreasing on the right of the critical point, the function will have a relative maximum at the critical point, we conclude that the function has a relative minimum at x = -2 and x = 4, and a relative maximum at x= 5 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9maalaaabaGaaGynaaqaaiaaikdaaaaaaa@397C@ .

 

13. f ′ (x)=4 x 3 −3 x 2 −6x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaaGinaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiAdacaWG4baaaa@4334@ . So, f ″ (x)=12 x 2 −6x−6=6(2 x 2 −x−1)=6(x−1)(2x+1) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaagaGaaiikaiaadIhacaGGPaGaeyypa0JaaGymaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOnaiaadIhacqGHsislcaaI2aGaeyypa0JaaGOnaiaacIcacaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadIhacqGHsislcaaIXaGaaiykaiabg2da9iaaiAdacaGGOaGaamiEaiabgkHiTiaaigdacaGGPaGaaiikaiaaikdacaWG4bGaey4kaSIaaGymaiaacMcaaaa@55D0@ .

The graph of the function is concave up where the second order derivative f ″ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaagaGaaiikaiaadIhacaGGPaaaaa@393C@  is positive and is concave down where the second order derivative is negative. So, we have to find the sign graph of the second order derivative f ″ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaagaGaaiikaiaadIhacaGGPaaaaa@393C@ . The second order derivative is zero when x = 1 and x= −1 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9maalaaabaGaeyOeI0IaaGymaaqaaiaaikdaaaaaaa@3A65@ . These 2 points divide the real line into 3 intervals in each of which the second order derivative has the same sign. We pick points and test the value to find the signs, or we determine the sign in one interval and consider the exponent of the linear factors to find the signs in rest of the interval. Below is the sign graph of the second order derivative f ″ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaagaGaaiikaiaadIhacaGGPaaaaa@393C@

 

Thus, the function is concave up on ( −∞, −1 2 )∪( 1,∞ ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacqGHsislcqGHEisPcaGGSaWaaSaaaeaacqGHsislcaaIXaaabaGaaGOmaaaaaiaawIcacaGLPaaacqWIQisvdaqadaqaaiaaigdacaGGSaGaeyOhIukacaGLOaGaayzkaaaaaa@4290@  and concave down on ( −1 2 ,1 ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkHiTiaaigdaaeaacaaIYaaaaiaacYcacaaIXaaacaGLOaGaayzkaaaaaa@3B56@ . Since the concavity is changing at x= −1 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9maalaaabaGaeyOeI0IaaGymaaqaaiaaikdaaaaaaa@3A65@  and at x = 1, there are points of inflections at these two points.

 

14. f(x)= x 3 −6 x 2 +9x=x( x 2 −6x+9)=x (x−3) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaI2aGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiMdacaWG4bGaeyypa0JaamiEaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOnaiaadIhacqGHRaWkcaaI5aGaaiykaiabg2da9iaadIhacaGGOaGaamiEaiabgkHiTiaaiodacaGGPaWaaWbaaSqabeaacaaIYaaaaaaa@52EC@ . Below is the sign graph of f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@ .

f ′ (x)=3 x 2 −12x+9=3( x 2 −4x+3)=3(x−1)(x−3) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIXaGaaGOmaiaadIhacqGHRaWkcaaI5aGaeyypa0JaaG4maiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGinaiaadIhacqGHRaWkcaaIZaGaaiykaiabg2da9iaaiodacaGGOaGaamiEaiabgkHiTiaaigdacaGGPaGaaiikaiaadIhacqGHsislcaaIZaGaaiykaaaa@5508@ . Below is the sign graph of f ′ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaaaaa@393B@ .

From above, we can tell that f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@  has a relative maximum at x = 1 and relative minimum at x = 3.

f ″ (x)=6x−12=6(x−2) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaagaGaaiikaiaadIhacaGGPaGaeyypa0JaaGOnaiaadIhacqGHsislcaaIXaGaaGOmaiabg2da9iaaiAdacaGGOaGaamiEaiabgkHiTiaaikdacaGGPaaaaa@4428@ . Below is the sign graph of f ″ (x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaagaGaaiikaiaadIhacaGGPaaaaa@393C@

From above, we can tell that there is a point of inflection at x = 2.

Below is the graph of f(x) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaaaa@392F@ .

 

15. q=5000−100p MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyCaiabg2da9iaaiwdacaaIWaGaaGimaiaaicdacqGHsislcaaIXaGaaGimaiaaicdacaWGWbaaaa@3EE8@  or, 100p=5000−q MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaicdacaaIWaGaamiCaiabg2da9iaaiwdacaaIWaGaaGimaiaaicdacqGHsislcaWGXbaaaa@3EE8@  or, p=50− q 100 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiCaiabg2da9iaaiwdacaaIWaGaeyOeI0YaaSaaaeaacaWGXbaabaGaaGymaiaaicdacaaIWaaaaaaa@3D84@ .

Revenue = (quantity sold).(price per unit) = qp=q(50− q 100 )=50q− q 2 100 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyCaiaadchacqGH9aqpcaWGXbGaaiikaiaaiwdacaaIWaGaeyOeI0YaaSaaaeaacaWGXbaabaGaaGymaiaaicdacaaIWaaaaiaacMcacqGH9aqpcaaI1aGaaGimaiaadghacqGHsisldaWcaaqaaiaadghadaahaaWcbeqaaiaaikdaaaaakeaacaaIXaGaaGimaiaaicdaaaaaaa@4953@

Profit = revenue – MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqefmuySLMyYLgaiuaajugybabaaaaaaaaapeGaa83eGaaa@3A6A@  cost = 50q− q 2 100 −(3000−20q+0.3 q 2 )=50q− q 2 100 −3000+20q−0.3 q 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGynaiaaicdacaWGXbGaeyOeI0YaaSaaaeaacaWGXbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGymaiaaicdacaaIWaaaaiabgkHiTiaacIcacaaIZaGaaGimaiaaicdacaaIWaGaeyOeI0IaaGOmaiaaicdacaWGXbGaey4kaSIaaGimaiaac6cacaaIZaGaamyCamaaCaaaleqabaGaaGOmaaaakiaacMcacqGH9aqpcaaI1aGaaGimaiaadghacqGHsisldaWcaaqaaiaadghadaahaaWcbeqaaiaaikdaaaaakeaacaaIXaGaaGimaiaaicdaaaGaeyOeI0IaaG4maiaaicdacaaIWaGaaGimaiabgUcaRiaaikdacaaIWaGaamyCaiabgkHiTiaaicdacaGGUaGaaG4maiaadghadaahaaWcbeqaaiaaikdaaaaaaa@5F95@

= −0.01 q 2 +70q−3000−0.3 q 2 =−0.31 q 2 +70q−3000 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaGimaiaac6cacaaIWaGaaGymaiaadghadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI3aGaaGimaiaadghacqGHsislcaaIZaGaaGimaiaaicdacaaIWaGaeyOeI0IaaGimaiaac6cacaaIZaGaamyCamaaCaaaleqabaGaaGOmaaaakiabg2da9iabgkHiTiaaicdacaGGUaGaaG4maiaaigdacaWGXbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4naiaaicdacaWGXbGaeyOeI0IaaG4maiaaicdacaaIWaGaaGimaaaa@55BA@

Thus we have a profit function given by P(q)=−0.31 q 2 +70q−3000 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiaacIcacaWGXbGaaiykaiabg2da9iabgkHiTiaaicdacaGGUaGaaG4maiaaigdacaWGXbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4naiaaicdacaWGXbGaeyOeI0IaaG4maiaaicdacaaIWaGaaGimaaaa@46FD@  which depends on the number of units q sold.

The marginal profit is then P ′ (q)=−0.62q+70 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmiuayaafaGaaiikaiaadghacaGGPaGaeyypa0JaeyOeI0IaaGimaiaac6cacaaI2aGaaGOmaiaadghacqGHRaWkcaaI3aGaaGimaaaa@414C@  and the marginal profit when the production level is 500 units = P ′ (500)=−0.62(500)+70=−310+70=−240 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmiuayaafaGaaiikaiaaiwdacaaIWaGaaGimaiaacMcacqGH9aqpcqGHsislcaaIWaGaaiOlaiaaiAdacaaIYaGaaiikaiaaiwdacaaIWaGaaGimaiaacMcacqGHRaWkcaaI3aGaaGimaiabg2da9iabgkHiTiaaiodacaaIXaGaaGimaiabgUcaRiaaiEdacaaIWaGaeyypa0JaeyOeI0IaaGOmaiaaisdacaaIWaaaaa@4FC8@