Question

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.)

$f(x)=(11 - x)(x + 1)^2$

(a) find the critical numbers of $f$. (enter your answers as a comma-separated list.)

$x =$

(b) find the open intervals on which the function is increasing or decreasing. (enter your answers using interval notation.)

increasing

decreasing

(c) apply the first derivative test to identify all relative extrema.

relative maximum $(x,y)=(quad)$

relative minimum $(x,y)=(quad)$

Explanation:

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

Use the product rule \( (uv)^\prime = u^\prime v+uv^\prime \), where \( u = 11 - x \), \( u^\prime=-1 \), \( v=(x + 1)^2=x^{2}+2x + 1 \), \( v^\prime=2(x + 1)=2x+2 \).

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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
• 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
• At \( x = 7 \): \( f(7)=(11 - 7)(7 + 1)^2=4\times64 = 256 \). Since \( f(x) \) changes from increasing (\( -17 \)), \( (7,256) \) 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