Question

1)Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$f\left(x\right)=x^2$

2)Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$g\left(x\right)=x^3-3x$

3) Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$h\left(x\right)=\sin x+\cos x,0<x<\frac{\pi }{2}$

4) Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$f\left(x\right)=\sin x-\cos x,0<x<2\pi$

5) Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$f\left(x\right)=x^3-6x^2+9x+15$

6) Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$g\left(x\right)=\frac{x}{2}+\frac{2}{x},x>0$

7) Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$g\left(x\right)=\frac{1}{x^2+2}$

8)4) Find the local maxima and local minima, if any, of the function. Find, also the local maximum & the local minimum values, as the case may be:
$f\left(x\right)=x\sqrt{1-x},0<x<1$

Answer 1

1) Given: $f\left(x\right)=x^2$

Differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=2x$

Putting $f^{\prime }\left(x\right)=0$

$2x=0\Rightarrow x=0$

Again, Differentiating w.r.t $x$

$f^{\prime \prime }\left(x\right)=2>0$

As $f^{\prime \prime }\left(x\right)>0,$
so $x=0$ is point of local minima and there is no point of local maxima.

We know, $x=0$ is point of local minima, so the value of local minima is

$f\left(0\right)=0^2=0$

Thus, minimum value of $f\left(x\right)$ is $0$ and there is no maximum value.

2) Given: $g\left(x\right)=x^3-3x$

Differentiating w.r.t $x,$

$g^{\prime }\left(x\right)=3x^2-3$

Putting $g^{\prime }\left(x\right)=0$

$3x^2-3=0$

$\Rightarrow 3x^2=3$

$\Rightarrow x^2=\frac{3}{3}=1$

$\Rightarrow x=\pm 1$

Again, Differentiating w.r.t $x$

$g^{\prime \prime }\left(x\right)=6x-0$

$=6x$

When $x=-1$,

$g^{\prime \prime }\left(x\right)=-6<0$

So, $x=-1$ is point of local maxima

When $x=1$,

$g^{\prime \prime }\left(x\right)=6>0$

So, $x=1$ is point of local minima

Thus, $x=-1$ is point of local maxima and
$g\left(x\right)$ has local minima at $x=1$
Local minimum value at $x=1$

$g\left(x\right)=x^3-3x$

$g\left(1\right)=(1{)}^3-3(1)=-2$

Local maximum value at $x=-1$

$g\left(x\right)=x^3-3x$

$g(-1)=(-1{)}^3-3(-1)=2$

Thus, the Local maximum value of $g\left(x\right)$ is $2$ and Local
minimum value of $g\left(x\right)$ is $-2$.

3) $h\left(x\right)=\sin x+\cos x,0<x<\frac{\pi }{2}$

Differentiating w.r.t $x,$

$h^{\prime }\left(x\right)=\cos x-\sin x$

Putting $h^{\prime }\left(x\right)=0$

$\cos x-\sin x=0$

$\Rightarrow \cos x=\sin x$

$\Rightarrow \tan x=1$

$\Rightarrow \tan x=1$

$\Rightarrow x=\frac{\pi }{4}[\because 0<x<\frac{\pi }{2}]$

Again, Differentiating w.r.t $x,$

$h^{\prime \prime }\left(x\right)=-\sin x-\cos x$

When $x=\frac{\pi }{4}:$

$h^{\prime \prime }\left(\frac{\pi }{4}\right)=-\sin \left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{4}\right)$

$\Rightarrow h^{\prime \prime }\left(\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$

$\Rightarrow h^{\prime \prime }\left(\frac{\pi }{4}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}$

Since $h^{\prime \prime }\left(\frac{\pi }{4}\right)<0:$

so, $x=\frac{\pi }{4}$ is a point of local maxima
Local maximum value at $x=\frac{\pi }{4}$ is :
$f\left(x\right)=\sin x+\cos x$

$\Rightarrow f\left(\frac{\pi }{4}\right)=\sin \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}\right)$

$\Rightarrow f\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$

$\Rightarrow f\left(\frac{\pi }{4}\right)=\frac{2}{\sqrt{2}}$

$\Rightarrow f\left(\frac{\pi }{4}\right)=\sqrt{2}$

Thus, $x=\frac{\pi }{4}$ is point of local maxima and the maximum value of $h\left(x\right)$ is $\sqrt{2}.$

4) $f\left(x\right)=\sin x-\cos x,0<x<2\pi$

Differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=\cos x-(-\sin x)$

$\Rightarrow f^{\prime }\left(x\right)=\cos x+\sin x$

Putting $f^{\prime }\left(x\right)=0:$

$\cos x+\sin x=0$

$\Rightarrow \sin x=-\cos x$

$\Rightarrow \tan x=-1$

$\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}[\because 0,x<2\pi ]$

Again, Differentiating w.r.t $x,$

$f^{\prime \prime }\left(x\right)=-\sin x+\cos x$

When $x=\frac{3\pi }{4}:$
then $f^{\prime \prime }\left(\frac{3\pi }{4}\right)=-\sin \left(\frac{3\pi }{4}\right)+\cos \left(\frac{3\pi }{4}\right)$

$\Rightarrow f^{\prime \prime }\left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$

$f^{\prime \prime }\left(\frac{3\pi }{4}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}$

As $f^{\prime \prime }\left(x\right)<0,$ When $x=\frac{3\pi }{4}$

Thus $x=\frac{3\pi }{4}$ is point of local maxima.
When $x=\frac{7\pi }{4}:$
Then $f^{\prime \prime }\left(\frac{7\pi }{4}\right)=-\sin \left(\frac{7\pi }{4}\right)+\cos \left(\frac{7\pi }{4}\right)$

$\Rightarrow f^{\prime \prime }\left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$

$\Rightarrow f^{\prime \prime }\left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}}=\sqrt{2}$

As $f^{\prime \prime }\left(x\right)>0,$ when $x=\frac{7\pi }{4}$ is point of local minima.
The local maximum value of
$f\left(x\right)$ at $x=\frac{3\pi }{4}$ is
$f\left(x\right)=\sin x-\cos x$

$f\left(\frac{3\pi }{4}\right)=\sin \left(\frac{3\pi }{4}\right)-\cos \left(\frac{3\pi }{4}\right)$

$\Rightarrow f\left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$

$\Rightarrow f\left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}}=\sqrt{2}$

The local minimum value of $f\left(x\right)$ at $x=\frac{7\pi }{4}$ is

$f\left(x\right)=\sin x-\cos x$

$f\left(\frac{7\pi }{4}\right)=\sin \left(\frac{7\pi }{4}\right)-\cos \left(\frac{7\pi }{4}\right)$

$\Rightarrow f\left(\frac{7\pi }{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$

$\Rightarrow f\left(\frac{7\pi }{4}\right)=-\frac{1}{\sqrt{2}}=-\sqrt{2}$

Hence, the Local maximum value of $f\left(x\right)$ is $\sqrt{2}$ at$x=\frac{3\pi }{4}$
and the Local minimum value of $f\left(x\right)$ is $-\sqrt{2}$ at

$x=\frac{7\pi }{4}$

(5) $f\left(x\right)=x^3-6x^2+9x+15$

Differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=3x^2-12x+9$

$\Rightarrow f^{\prime }\left(x\right)=3(x^2-4x+3)$

$\Rightarrow f^{\prime }\left(x\right)=3(x-3)(x-1)$

Putting $f^{\prime }\left(x\right)=0$

$3(x-3)(x-1)=0$

$\Rightarrow x=1,3$

Again differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=6x-12=6(x-2)$

When $x=1$:

$f^{\prime \prime }\left(1\right)=6(1-2)=-6<0$

As $f^{\prime \prime }\left(x\right)<0,$ when $x=1$

Thus $x=1$ is point of local maxima.

When $x=3$:

$f^{\prime \prime }\left(3\right)=6(3-2)=6>0$

As $f^{\prime \prime }\left(x\right)>0,$ when $x=3$

Thus $x=3$ is point of local minima.

Local maximum value of $f\left(x\right)$ at $x=1$

$f\left(x\right)=x^3-6x^2+9x+15$

$\Rightarrow f\left(1\right)=(1{)}^3-6(1{)}^2+9\left(1\right)+15$

$=1-6+9+15$

$=19$

Local minimum value of $f\left(x\right)$ at $x=3$

$\Rightarrow f\left(3\right)=(3{)}^3-6(3{)}^2+9\left(3\right)+15$

$=27-54+27+15$

$=15$

Hence, the Local maximum value of $f\left(x\right)$ is $19$ at $x=1$ and the Local minimum value of $f\left(x\right)$ is $15$ at $x=3$

(6) $g\left(x\right)=\frac{x}{2}+\frac{2}{x},x>0$

Differentiating w.r.t $x,$

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

Putting $g^{\prime }\left(x\right)=0$

$\frac{1}{2}-\frac{2}{x^2}=0$

$\Rightarrow \frac{x^2-4}{2x^2}=0$

$\Rightarrow x^2=4[\because x>0]$

$\Rightarrow x=2[\because x>0]$

Again differentiating w.r.t $x,$

$g^{\prime \prime }\left(x\right)=-2\left[\frac{d}{dx}\left(\frac{1}{x^2}\right)\right]$

$\Rightarrow g^{\prime \prime }\left(x\right)=-2[-\frac{2}{x^3}]=\frac{4}{x^3}$

When $x=2$:

$g^{\prime \prime }\left(x\right)=\frac{4}{2^3}>0$

As $g^{\prime \prime }\left(x\right)>0,$ when $x=2$

Thus $x=2$ is point of local minima.

Local minimum value of $g\left(x\right)$ at $x=2$ is:
$g\left(x\right)=\frac{x}{2}+\frac{2}{x}$

$\Rightarrow g\left(2\right)=\frac{2}{2}+\frac{2}{2}=2$

Hence, the Local minimum value of $g\left(x\right)$ is $2$ at $x=2$

(7) $g\left(x\right)=\frac{1}{x^2+2}$

Differentiating w.r.t $x,$

$g^{\prime }\left(x\right)=\frac{d}{dx}\left[\right(x^2+2{)}^{-1}]$

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

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

Putting $g^{\prime }\left(x\right)=0$

$\frac{-2x}{(x^2+2{)}^2}=0$

$\Rightarrow x=0$

Using first derivative test, we get

$x<0\Rightarrow g^{\prime }\left(x\right)>0$

$x>0\Rightarrow g^{\prime }\left(x\right)<0$

So, $x=0$ is point of local maxima.
Local maximum value of $g\left(x\right)$ at $x=0$ is

$g\left(x\right)=\frac{1}{x^2+2}$

$\Rightarrow g\left(0\right)=\frac{1}{0^2+2}=\frac{1}{2}$

Hence, the Local maximum value of $g\left(x\right)$ is $\frac{1}{2}$

​​​​at $x=0$

(8) $f\left(x\right)=x\sqrt{1-x},0<x<1$

Differentiation w.r.t $x,$

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(x\sqrt{1-x}\right)$

$\Rightarrow f^{\prime }\left(x\right)=x\left(\frac{-1}{2\sqrt{1-x}}\right)+\sqrt{1-x}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-x}{2\sqrt{1-x}}+\sqrt{1-x}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-x+2(1-x)}{2\sqrt{1-x}}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{2-3x}{2\sqrt{1-x}}$

Putting $f^{\prime }\left(x\right)=0:$

$\frac{2-3x}{2\sqrt{1-x}}=0$

$\Rightarrow 2-3x=0[0<x<1]$

$\Rightarrow x=\frac{2}{3}$

Using first derivative test, we get

$x<\frac{2}{3}\Rightarrow f^{\prime }\left(x\right)>0$

$x>\frac{2}{3}\Rightarrow f^{\prime }\left(x\right)<0$

So, $x=\frac{2}{3}$ is point of local maxima.
Local maximum value of $f\left(x\right)$ at
$x=\frac{2}{3}$ is :
$f\left(x\right)=x\sqrt{1-x}$

$\Rightarrow f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}$

$\Rightarrow f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{\frac{1}{3}}$

$\Rightarrow f\left(\frac{2}{3}\right)=\frac{2}{3\sqrt{3}}$

Hence, the Local maximum value of $f\left(x\right)$ is

$\frac{2}{3\sqrt{3}}$ at $x=\frac{2}{3}$