consider the following function. use a graphing utility to confirm your answers for parts (a) through (c)…

consider the following function. use a graphing utility to confirm your answers for parts (a) through (c). (if an answer does not exist, enter dne.)\n\n$f(x)=(11 - x)(x + 1)^2$\n\n(a) find the critical numbers of $f$. (enter your answers as a comma-separated list.)\n\n$x =$\n\n(b) find the open intervals on which the function is increasing or decreasing. (enter your answers using interval notation.)\n\nincreasing\n\ndecreasing\n\n(c) apply the first derivative test to identify all relative extrema.\n\nrelative maximum $(x,y)=(quad)$\n\nrelative minimum $(x,y)=(quad)$

consider the following function. use a graphing utility to confirm your answers for parts (a) through (c). (if an answer does not exist, enter dne.)\n\n$f(x)=(11 - x)(x + 1)^2$\n\n(a) find the critical numbers of $f$. (enter your answers as a comma-separated list.)\n\n$x =$\n\n(b) find the open intervals on which the function is increasing or decreasing. (enter your answers using interval notation.)\n\nincreasing\n\ndecreasing\n\n(c) apply the first derivative test to identify all relative extrema.\n\nrelative maximum $(x,y)=(quad)$\n\nrelative minimum $(x,y)=(quad)$

Answer

Explanation:

Step1: Find the derivative of ( f(x) )

Use the product rule ( (uv)^\prime = u^\prime v+uv^\prime ), where ( u = 11 - x ), ( u^\prime=-1 ), ( v=(x + 1)^2=x^{2}+2x + 1 ), ( v^\prime=2(x + 1)=2x+2 ). [ \begin{align*} f^\prime(x)&=-1\times(x + 1)^2+(11 - x)\times2(x + 1)\ &=(x + 1)[- (x + 1)+2(11 - x)]\ &=(x + 1)(-x-1 + 22-2x)\ &=(x + 1)(21-3x)\ &=-3(x + 1)(x - 7) \end{align*} ]

Step2: Find the critical numbers

Set ( f^\prime(x)=0 ), then ( -3(x + 1)(x - 7)=0 ). Solving ( x+1 = 0 ) gives ( x=-1 ), solving ( x - 7=0 ) gives ( x = 7 ).

Step3: Determine the intervals of increase and decrease

  • For ( x<-1 ), let ( x=-2 ), then ( f^\prime(-2)=-3(-2 + 1)(-2 - 7)=-27<0 ), so ( f(x) ) is decreasing on ( (-\infty,-1) ).
  • For ( -1<x<7 ), let ( x = 0 ), then ( f^\prime(0)=-3(0 + 1)(0 - 7)=21>0 ), so ( f(x) ) is increasing on ( (-1,7) ).
  • For ( x>7 ), let ( x = 8 ), then ( f^\prime(8)=-3(8 + 1)(8 - 7)=-27<0 ), so ( f(x) ) is decreasing on ( (7,\infty) ).

Step4: Apply the First - Derivative Test

  • At ( x=-1 ): ( f(x)=(11+1)(-1 + 1)^2=0 ). Since ( f(x) ) changes from decreasing (( x<-1 )) to increasing (( -1<x<7 )), this is not a relative extremum (it's a point where the slope changes direction but the function value is ( 0 ) and the behavior is not a local max/min in the traditional sense for non - zero function values in the neighborhood).
  • At ( x = 7 ): ( f(7)=(11 - 7)(7 + 1)^2=4\times64 = 256 ). Since ( f(x) ) changes from increasing (( -1<x<7 )) to decreasing (( x>7 )), ( (7,256) ) is a relative maximum.

Answer:

(a) ( x=-1,7 ) (b) Increasing: ( (-1,7) ); Decreasing: ( (-\infty,-1)\cup(7,\infty) ) (c) Relative maximum: ( (7,256) ); Relative minimum: DNE