Question

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

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

Answer 1

The correct option is B ${\left(e\right)}^{1/e}$

For every real number (or) valued function f(x), the values of

x which satisfies the equation $f^1\left(x\right)=0$ are the point of

it's local and global maxima or minima.

This occus due to the fact that, at the point of

maxima or minima, the curve of the function has a

zero slope.

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

We will be using the equation,

$y={\left(1/x\right)}^x$

Taking in both sides we get

$lny=-xlnx$

Differentiating both sides with respect to x.

$y.\frac{dy}{dx}=-lnx-1$

$\frac{dy}{dx}=-y(lnx+1)$

Equating $\frac{dy}{dx}$ to 0, we get

$-y(lnx+1)=0$

Since y is an exponential function it can never be equal

to zero, hence

$lnx+1=0$

$lnx=-1$

$x=e^{-1}$

So, for the maximum value we put $x=e^{-1}$ in f(x) to

get the value of f(x) at the point.

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

Hence the maximum value of the function is

$e^{1/e}$.

$\therefore$ So, the answer is $B.e^{1/e}$