Question

How do you find the local max and min for ${{x}^{5}}\ln x$?

Answer 1

Hint: In this question we have been given an expression for which we have to find the local maximum and the minimum value. To find the values we will consider the function as $f\left( x \right)$ and then find the derivative of the function and write it as $f'\left( x \right)$. Now to find the extremes, we will substitute $f'\left( x \right)=0$ and find the values of $x$. We will check whether $f'\left( x \right)$ is greater or lesser than $0$ and find the extreme values.

Complete step-by-step solution:
Consider the expression given to be $f\left( x \right)$ therefore, it can be written as:
$\Rightarrow f\left( x \right)={{x}^{5}}\ln x$
Now to find the extremes, we will take the derivative of $f\left( x \right)$. We can write it as:
$\Rightarrow f'\left( x \right)={{x}^{5}}\dfrac{d}{dx}\ln x+\ln x\dfrac{d}{dx}{{x}^{5}}$, using the product rule.
Now we know that $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ therefore, on substituting, we get:
$\Rightarrow f'\left( x \right)={{x}^{5}}\times \dfrac{1}{x}+\ln x\times 5{{x}^{4}}$
On simplifying, we get:
$\Rightarrow f'\left( x \right)={{x}^{4}}+5{{x}^{4}}\ln x$
Now we will find the critical points by considering $f'\left( x \right)=0$.
$\Rightarrow {{x}^{4}}+5{{x}^{4}}\ln x=0$
On taking ${{x}^{4}}$ common, we get:
$\Rightarrow {{x}^{4}}\left( 1+5\ln x \right)=0$
Therefore, we get:
$\Rightarrow {{x}^{4}}=0$ and $1+5\ln x=0$
On simplifying, we get:
$\Rightarrow x=0$ and $x={{e}^{-\dfrac{1}{5}}}$
Now we know that $\ln 0$, which is the natural log of $0$ is undefined for $x=0$ therefore, the only feasible solution will be $1+5\ln x=0$.
Now we will evaluate the inequality $f'\left( x \right)>0$.
Since the term ${{x}^{4}}>0$ for all $x>0$, this indicates that:
$\Rightarrow f'\left( x \right)>0$ for $x\in \left( {{e}^{-\dfrac{1}{5}}},+\infty \right)$ and $f'\left( x \right)<0$ for $x\in \left( 0,{{e}^{-\dfrac{1}{5}}} \right)$
This shows that the function is decreasing in the range $\left( 0,{{e}^{-\dfrac{1}{5}}} \right)$ and increasing in the range $x\in \left( {{e}^{-\dfrac{1}{5}}},+\infty \right)$ which means that:
$\Rightarrow x={{e}^{-\dfrac{1}{5}}}$, is the local minimum. Which is the required solution.

Note: It is to be remembered that maximum and minimum values of a function are called as the extreme values and the concept of derivatives is used to find the extreme values. The logarithm used in the question is the natural log which has the base $e$. It is to be noted that the local maximum of the given function is not a specific value. It has the value of $\infty $.