Find the exact location of all the relative and absolute extrema of the function with domain $( 0 , \infty )$ . $f ( x ) = x \ln \left( x ^ { 3 } \right)$
A) $\left( \frac { 1 } { e } , \frac { 3 } { e } \right)$ - absolute minimum
B) $\left( \frac { 1 } { e } , - \frac { 3 } { e } \right)$ - relative maximum
C) $\left( \frac { 3 } { e } , - \frac { 1 } { 3 e } \right)$ - absolute minimum
D) $\left( \frac { 1 } { e } , - \frac { 1 } { 3 e } \right)$ - relative minimum
E) $\left( \frac { 1 } { e } , - \frac { 3 } { e } \right)$ - absolute minimum

Question

Find the exact location of all the relative and absolute extrema of the function with domain $( 0 , \infty )$ .   $f ( x ) = x \ln \left( x ^ { 3 } \right)$  
A) $\left( \frac { 1 } { e } , \frac { 3 } { e } \right)$ - absolute minimum
B) $\left( \frac { 1 } { e } , - \frac { 3 } { e } \right)$ - relative maximum
C) $\left( \frac { 3 } { e } , - \frac { 1 } { 3 e } \right)$ - absolute minimum
D) $\left( \frac { 1 } { e } , - \frac { 1 } { 3 e } \right)$ - relative minimum
E) $\left( \frac { 1 } { e } , - \frac { 3 } { e } \right)$ - absolute minimum

Quizplus · Accepted Answer

Explanation:$f'(x) = \ln(x^3) + 3 = 0$$\Rightarrow \ln(x^3) = -3$$\Rightarrow x^3 = e^{-3}$$\Rightarrow x = \frac{1}{e}$$f''(x) = \frac{3}{x} > 0$ for $x > 0$, so $x = \frac{1}{e}$ is an absolute minimum.

Find the Exact Location of All the Relative and Absolute (0,∞)( 0 , \infty )(0,∞)

Find the Exact Location of All the Relative and Absolute $( 0 , \infty )$