Question

Let $exp\left(x\right)$ denote the exponential function $e^x$. If $f\left(x\right)=exp\left(x^{\frac{1}{x}}\right)$, $x>0$, then the minimum value of $f$ in the interval $[2,5]$ is

• A. $exp\left(e^{\frac{1}{e}}\right)$
• B. $exp\left(2^{\frac{1}{2}}\right)$
• C. $exp\left(5^{\frac{1}{5}}\right)$
• D. $exp\left(3^{\frac{1}{3}}\right)$

Answer 1

Answer: (C)

$exp\left(5^{\frac{1}{5}}\right)$

Given that $e^{{\left(x\right)}^{\frac{1}{x}}}$, $x>0$,
Taking log on both sides, we get
$\log +x={\left(x\right)}^{\frac{1}{x}}=g\left(x\right)\left(say\right)....\left(i\right)$
Here, $g\left(x\right)={\left(x\right)}^{\frac{1}{x}}$
$\Rightarrow \log g\left(x\right)=\frac{1}{x}\log x$
On differentiating w.r.t $x$ we get
$\frac{1}{g\left(x\right)}.g^{\prime }\left(x\right)=\frac{x.\frac{1}{x}-\log x}{x^2}$ $=\left(\frac{1-\log x}{x^2}\right)$

$\Rightarrow g^{\prime }\left(x\right)={\left(x\right)}^{(\frac{1}{x}-2)}(1-\log x)$

For maximum or minimum of $g\left(x\right)$ put
$g^{\prime }\left(x\right)=0$
$\Rightarrow {\left(x\right)}^{(\frac{1}{x}-2)}(1-\log x)=0$

$\Rightarrow \log x=1=\log e$
and

$g^{\prime \prime }\left(x\right){|}_{x=e}>0$

So, $g\left(x\right)$ is minimum at $x=e$

$\therefore$ $\left(g\right(x)$ in $(0,e)$ and decreases in $(e,-\infty )$, it will be minimum at either $2$ or $5$

$\therefore 2^{\frac{1}{2}}>5^{\frac{1}{5}}$ $\Rightarrow$ Minimum value of $f\left(x\right)=e^{{\left(5\right)}^{\frac{1}{5}}}$