Question

Let $exp\left(x\right)$ denote 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)

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

$f\left(x\right)=e^{x^{1/x}},x>0,xϵ[2,5]$

$f\left(2\right)=e^{2^{1/2}}$

$f\left(3\right)=e^{3^{1/3}}$

$f\left(4\right)=e^{4^{1/4}}$

$f\left(5\right)=e^{5^{1/5}}$

We know that

$e^{5^{1/5}}<e^{4^{1/4}}<e^{3^{1/3}}<e^{2^{1/2}}$

$\therefore f\left(5\right)$ will have minimum value, equal to exp $\left(5^{1/5}\right)$