Question

Consider the function
$f\left(x\right)={\left(\frac{1}{x}\right)}^{2x^2}$, where $x>0$
At what value of $x$ does the function attain maximum value?

• A. $e$
• B. $\sqrt{e}$
• C. $\frac{1}{\sqrt{e}}$
• D. $\frac{1}{e}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{e}}$

Let $y=f\left(x\right)={\left(\frac{1}{x}\right)}^{2x^2}$

Taking natural Logaritm on both sides, we have

$\ln y=-2x^2\ln x$

Now, differentiating both sides wrt $x$, we have $\frac{1}{y}\frac{dy}{dx}=-4x\ln x-2x$

$⟹\frac{dy}{dx}=(-4x\ln x-2x){\left(\frac{1}{x}\right)}^{2x^2}=0⟹(-4x\ln x-2x)=0⟹x=e^{-0.5}$

Again differentiating wrt $x$, we have

$\frac{d^{2}y}{d{x}^2}=(-4x\ln x-2x{)}^2{\left(\frac{1}{x}\right)}^{2x^2}+(-4\ln x-6)$

at $e^{-0.5}$, $\frac{d^{2}y}{d{x}^2}=(2e^{-0.5}-2e^{-0.5}{)}^2{\left(e^{0.5}\right)}^{2e^{-1}}+(2-6)=-4<0$

Hence, there is a maxima at $x=e^{-0.5}$.

This is the required answer.