Question

find the absolute maximum value of $f(x)=3x^{5}lnleft(\frac{1}{x}
ight)$ and say where it is assumed.
the absolute maximum is $square$ at $x = square$.
(type exact answers.)

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=3x^{5}\ln(\frac{1}{x})$ as $f(x)= - 3x^{5}\ln(x)$ with domain $x>0$.

Step2: Find the derivative

Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u=-3x^{5}$ and $v = \ln(x)$. $u^\prime=-15x^{4}$ and $v^\prime=\frac{1}{x}$. Then $f^\prime(x)=-15x^{4}\ln(x)-3x^{5}\cdot\frac{1}{x}=-15x^{4}\ln(x)-3x^{4}=-3x^{4}(5\ln(x) + 1)$.

Step3: Find the critical points

Set $f^\prime(x) = 0$. Since $x^{4}>0$ for $x>0$, we solve $5\ln(x)+1 = 0$. $\ln(x)=-\frac{1}{5}$, so $x = e^{-\frac{1}{5}}$.

Step4: Determine if it's a maximum

Take the second - derivative of $f(x)$. Using the product - rule on $f^\prime(x)=-15x^{4}\ln(x)-3x^{4}$. Let $u=-15x^{4}$ and $v=\ln(x)$ for the first part of $f^\prime(x)$. $u^\prime=-60x^{3}$ and $v^\prime=\frac{1}{x}$, then the derivative of $-15x^{4}\ln(x)$ is $-60x^{3}\ln(x)-15x^{3}$, and the derivative of $-3x^{4}$ is $-12x^{3}$. So $f^{\prime\prime}(x)=-60x^{3}\ln(x)-15x^{3}-12x^{3}=-60x^{3}\ln(x)-27x^{3}$. Evaluate $f^{\prime\prime}(e^{-\frac{1}{5}})=-60(e^{-\frac{1}{5}})^{3}\cdot(-\frac{1}{5})-27(e^{-\frac{1}{5}})^{3}=12(e^{-\frac{3}{5}})-27(e^{-\frac{3}{5}})=- 15e^{-\frac{3}{5}}<0$. So $x = e^{-\frac{1}{5}}$ is a local maximum.

Step5: Find the maximum value

Substitute $x = e^{-\frac{1}{5}}$ into $f(x)$: $f(e^{-\frac{1}{5}})=-3(e^{-\frac{1}{5}})^{5}\ln(e^{-\frac{1}{5}})=-3e^{-1}\cdot(-\frac{1}{5})=\frac{3}{5e}$.

Answer:

The absolute maximum is $\frac{3}{5e}$ at $x = e^{-\frac{1}{5}}$