Question

The function $f\left(x\right)=x^{\frac{1}{x}}$ has

• A. Increasing in $\left(1,\infty \right)$
• B. Decreasing in $\left(1,\infty \right)$
• C. Increasing in $\left(1,e\right)$ decreasing in $\left(e,\infty \right)$
• D. Decreasing in $\left(1,e\right)$ increasing in $\left(e,\infty \right)$

Answer 1

The correct option is C

Increasing in $\left(1,e\right)$ decreasing in $\left(e,\infty \right)$

Find the interval of increase and decrease for the given function:

Given function, $f\left(x\right)=x^{\frac{1}{x}}$

Let $y=f\left(x\right)$

$\Rightarrow \log \left(y\right)=\frac{1}{x}\log \left(x\right)$

Now differentiating with respect to $x$ we get

$$\begin{gathered} \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-\frac{1}{x^2}\log \left(x\right) \\ \Rightarrow \frac{dy}{dx}=y\left(\frac{1-\log \left(x\right)}{x^2}\right) \\ \Rightarrow \frac{dy}{dx}=x^{\frac{1}{x}}\left(\frac{1-\log \left(x\right)}{x^2}\right) \end{gathered}$$

$x^{\frac{1}{x}}>0,\left\{\frac{\left[1-\log \left(x\right)\right]}{x^2}\right\}>0$in $\left(1,e\right)$ and $\left\{\frac{\left[1-\log \left(x\right)\right]}{x^2}\right\}<0$ in $\left(e,\infty \right)$

So, the function is increasing in $\left(1,e\right)$ and decreasing in $\left(e,\infty \right)$.

Hence, the correct option is $\left(\mathrm{C}\right)$.