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)=\frac{x^{9}-9 x}{9} )

(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 ( quad(x, y)=(quad) )

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

Explanation:

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

Given \( f(x)=\frac{x^{9}-9x}{9}=\frac{1}{9}x^{9}-x \).
Using the power rule \( (x^n)^\prime = nx^{n - 1} \), the derivative \( f^\prime(x)=\frac{1}{9}\times9x^{8}-1=x^{8}-1=(x^{4}+1)(x^{4}-1)=(x^{4}+1)(x^{2}+1)(x + 1)(x - 1) \).

Step2: Find the critical numbers

Set \( f^\prime(x)=0 \).
Since \( x^{4}+1>0 \) and \( x^{2}+1>0 \) for all real \( x \), then \( (x + 1)(x - 1)=0 \).
Solving \( (x + 1)(x - 1)=0 \) gives \( x=-1,x = 1 \).

Step3: Determine the intervals of increase and decrease

• Choose test points:
• For the interval \( (-\infty,-1) \), let \( x=-2 \). Then \( f^\prime(-2)=(-2)^{8}-1=256 - 1=255>0 \).
• For the interval \( (-1,1) \), let \( x = 0 \). Then \( f^\prime(0)=0^{8}-1=-1<0 \).
• For the interval \( (1,\infty) \), let \( x = 2 \). Then \( f^\prime(2)=2^{8}-1=256 - 1=255>0 \).
• The function \( f(x) \) is increasing when \( f^\prime(x)>0 \). So the increasing intervals are \( (-\infty,-1)\cup(1,\infty) \).
• The function \( f(x) \) is decreasing when \( f^\prime(x)<0 \). So the decreasing interval is \( (-1,1) \).

Step4: Apply the First - Derivative Test

• For \( x=-1 \):
• \( f(-1)=\frac{(-1)^{9}-9\times(-1)}{9}=\frac{-1 + 9}{9}=\frac{8}{9} \). Since \( f^\prime(x) \) changes from positive (left of \( x=-1 \)) to negative (right of \( x=-1 \)), \( (-1,\frac{8}{9}) \) is a relative maximum.
• For \( x = 1 \):
• \( f(1)=\frac{1^{9}-9\times1}{9}=\frac{1 - 9}{9}=-\frac{8}{9} \). Since \( f^\prime(x) \) changes from negative (left of \( x = 1 \)) to positive (right of \( x = 1 \)), \( (1,-\frac{8}{9}) \) is a relative minimum.

Answer:

(a) \( x=-1,1 \)
(b) Increasing: \( (-\infty,-1)\cup(1,\infty) \); Decreasing: \( (-1,1) \)
(c) Relative maximum: \( (-1,\frac{8}{9}) \); Relative minimum: \( (1,-\frac{8}{9}) \)