Question

Maximum value of ${\left(\frac{1}{x}\right)}^x$ is

• A. ${\left(e\right)}^e$
• B. ${\left(e\right)}^{\frac{1}{e}}$
• C. ${\left(e\right)}^{-e}$
• D. ${\left(\frac{1}{e}\right)}^e$

Answer 1

The correct option is B

${\left(e\right)}^{\frac{1}{e}}$

Explanation for the correct option:

Finding Maximum value of ${\left(\frac{1}{x}\right)}^x$ is

Given: ${\left(\frac{1}{x}\right)}^x$

We have function $f\left(x\right)={\left(\frac{1}{x}\right)}^x$

We will be using the equation, $y={\left(\frac{1}{x}\right)}^x$

Taking $\log$ both sides we get

$\ln y=-x\ln x$

Differentiating both sides w.r.t.$x$

$$\begin{gathered} \frac{1}{y}.\frac{dy}{dx}=-\ln x-1 \\ \Rightarrow \frac{dy}{dx}=-y(\ln x+1) \end{gathered}$$

Equating $\frac{dy}{dx}$ to $0$, we get $-y(\ln x+1)=0$

Since $y$ is an exponential function it can never be equal to zero,

Hence $\ln x+1=0$

$\Rightarrow$ $\ln x=-1$

$\Rightarrow$ $x=e^{-1}$

At neighbourhood of $x=e^{-1}$,$\frac{dy}{dx}$changes sign from positive to negative,hence maximum value exist at $x=e^{-1}$

So, for the maximum value we put $x=e^{-1}\text{in}f\left(x\right)$ to get the value of $f\left(x\right)$ at the point.

$f\left(e^{-1}\right)=e^{\frac{1}{e}}$

Hence the maximum value of the function is $e^{1/e}.$

Hence, the correct option is (B)