Given $f ( x ) = x + \frac { 36 } { x }$.
(a) Find the intervals on which $f$ is increasing or decreasing.
(b) Find the relative maxima and relative minima of $f$.

A) (a) Increasing on $( - 6,6 )$, decreasing on $( - \infty , - 6 )$ and $( 6 , \infty )$
(b) Rel. $\max . f ( 6 ) = 12$,
rel. min. $f ( - 6 ) = - 12$

B) (a) Increasing on $( - 6,0 )$ and $( 0,6 )$, decreasing on $( - \infty , - 6 )$ and $( 6 , \infty )$
(b) Rel. $\max . f ( 6 ) = 12$,
rel. min. $f ( - 6 ) = - 12$

C) (a) Increasing on $( - \infty , - 6 )$ and $( 6 , \infty )$, decreasing on $( - 6,0 )$ and $( 0,6 )$
(b) Rel. max. $f ( - 6 ) = - 12$,
rel. min. $f ( 6 ) = 12$

D) (a) Increasing on $( - \infty , - 6 )$ and $( 6 , \infty )$, decreasing on $( - 6,6 )$
(b) Rel. max. $f ( - 6 ) = - 12$,
rel. min. $f ( 6 ) = 12$

Question

Given $f ( x ) = x + \frac { 36 } { x }$.
(a) Find the intervals on which $f$ is increasing or decreasing.
(b) Find the relative maxima and relative minima of $f$.

A) (a) Increasing on $( - 6,6 )$, decreasing on $( - \infty , - 6 )$ and $( 6 , \infty )$
(b) Rel. $\max . f ( 6 ) = 12$,
rel. min. $f ( - 6 ) = - 12$

B) (a) Increasing on $( - 6,0 )$ and $( 0,6 )$, decreasing on $( - \infty , - 6 )$ and $( 6 , \infty )$
(b) Rel. $\max . f ( 6 ) = 12$,
rel. min. $f ( - 6 ) = - 12$

C) (a) Increasing on $( - \infty , - 6 )$ and $( 6 , \infty )$, decreasing on $( - 6,0 )$ and $( 0,6 )$
(b) Rel. max. $f ( - 6 ) = - 12$,
rel. min. $f ( 6 ) = 12$

D) (a) Increasing on $( - \infty , - 6 )$ and $( 6 , \infty )$, decreasing on $( - 6,6 )$
(b) Rel. max. $f ( - 6 ) = - 12$,
rel. min. $f ( 6 ) = 12$

Quizplus · Accepted Answer

Part (a) is correct, and all answer choices include the correct intervals for increasing and decreasing. 
For part (b), we can find the relative extrema by setting the derivative equal to zero and solving for $x$:

$$f'(x) = 1 - \frac{36}{x^2} = 0$$
$$\frac{36}{x^2} = 1$$
$$x^2 = 36$$
$$x = \pm 6$$

Therefore, we have relative extrema at $x = -6$ and $x = 6$. To determine whether these are relative maxima or minima, we can use the second derivative test:

$$f''(x) = \frac{72}{x^3}$$
At $x = -6$, $f''(-6) < 0$, so $x = -6$ is a relative maximum. At $x = 6$, $f''(6) > 0$, so $x = 6$ is a relative minimum.

Option (C) is the only choice that correctly identifies these relative extrema and their types.

Given f(x)=x+36xf ( x ) = x + \frac { 36 } { x }f(x)=x+x36​

Given $f ( x ) = x + \frac { 36 } { x }$