consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{3}-12x^{2}-27x +…

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{3}-12x^{2}-27x + 3$\n\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection point.\n\n$(x,y)=$\n\nfind the interval(s) on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval(s) on which $f$ is concave down. (enter your answer using interval notation.)

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{3}-12x^{2}-27x + 3$\n\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection point.\n\n$(x,y)=$\n\nfind the interval(s) on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval(s) on which $f$ is concave down. (enter your answer using interval notation.)

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ( (x^n)^\prime=nx^{n - 1} ), for ( f(x)=x^{3}-12x^{2}-27x + 3 ), we have ( f^\prime(x)=3x^{2}-24x-27=3(x^{2}-8x - 9)=3(x + 1)(x - 9) )

Step2: Determine critical points

Set ( f^\prime(x)=0 ), then ( 3(x + 1)(x - 9)=0 ). So ( x=-1 ) or ( x = 9 )

Step3: Test intervals for increasing/decreasing

  • For ( x<-1 ), let ( x=-2 ), ( f^\prime(-2)=3((-2)+1)((-2)-9)=3\times(-1)\times(-11)=33>0 )
  • For ( -1<x<9 ), let ( x = 0 ), ( f^\prime(0)=3(0 + 1)(0 - 9)=-27<0 )
  • For ( x>9 ), let ( x = 10 ), ( f^\prime(10)=3(10 + 1)(10 - 9)=33>0 )

So ( f(x) ) is increasing on ( (-\infty,-1)\cup(9,\infty) ) and decreasing on ( (-1,9) )

Step4: Find local extrema

Using the first - derivative test:

  • At ( x=-1 ), since ( f(x) ) changes from increasing to decreasing, ( f(-1)=(-1)^{3}-12(-1)^{2}-27(-1)+3=-1-12 + 27+3=17 ) (local maximum)
  • At ( x = 9 ), since ( f(x) ) changes from decreasing to increasing, ( f(9)=9^{3}-12\times9^{2}-27\times9+3=729-972-243 + 3=-483 ) (local minimum)

Step5: Find the second derivative

( f^\prime(x)=3x^{2}-24x-27 ), then ( f^{\prime\prime}(x)=6x-24=6(x - 4) )

Step6: Determine inflection point and concavity

Set ( f^{\prime\prime}(x)=0 ), then ( 6(x - 4)=0\Rightarrow x = 4 )

  • When ( x = 4 ), ( y=f(4)=4^{3}-12\times4^{2}-27\times4+3=64-192-108 + 3=-233 )
  • For ( x<4 ), let ( x = 0 ), ( f^{\prime\prime}(0)=6(0 - 4)=-24<0 ), so ( f(x) ) is concave down on ( (-\infty,4) )
  • For ( x>4 ), let ( x = 5 ), ( f^{\prime\prime}(5)=6(5 - 4)=6>0 ), so ( f(x) ) is concave up on ( (4,\infty) )

Answer:

(a) Increasing: ( (-\infty,-1)\cup(9,\infty) ); Decreasing: ( (-1,9) ) (b) Local minimum value: ( -483 ); Local maximum value: ( 17 ) (c) Inflection point: ( (4,-233) ); Concave up: ( (4,\infty) ); Concave down: ( (-\infty,4) )