Question

find local maxima and minima of f(x)=x√2-x^2​

Answer 1

Given,

$\longrightarrow f(x)=x\sqrt2-x^2$

We've to find its local maxima and minima.

Equating the first derivative of $f$ wrt $x$ to zero,

$\longrightarrow f'(x)=0$

$\longrightarrow \sqrt2-2x=0$

$\longrightarrow x=\dfrac{1}{\sqrt2}$

So $f$ is extremum at $x=\dfrac{1}{\sqrt2}.$

Let's check the nature of the second derivative of $f$ wrt $x$ at $x=\dfrac{1}{\sqrt2}.$

$\longrightarrow f''(x)=-2$

$\Longrightarrow f''\left(\dfrac{1}{\sqrt2}\right)<0\quad\!\forall x\in\mathbb{R}$

This means $f$ is maximum at $x=\dfrac{1}{\sqrt2}.$

$\longrightarrow f\left(\dfrac{1}{\sqrt2}\right)=\dfrac{1}{\sqrt2}\cdot\sqrt2-\left(\dfrac{1}{\sqrt2}\right)^2$

$\longrightarrow f\left(\dfrac{1}{\sqrt2}\right)=\dfrac{1}{2}$

So $f$ is maximum at $\left(\dfrac{1}{\sqrt2},\ \dfrac{1}{2}\right)$ and it has no minimum.