Question

1)Find the absolute maximum value and the absolute minimum value of the function in the given interval:
i) $f\left(x\right)=x^3,xϵ[-2,2]$

2)Find the absolute maximum value and the absolute minimum value of the function in the given interval:
ii) $f\left(x\right)=\sin x+\cos x,xϵ[0,\pi ]$

3)Find the absolute maximum value and the absolute minimum value of the function in the given interval:
iii) $f\left(x\right)=4x-\frac{1}{2}{x}^2,xϵ[-2,\frac{9}{2}]$

4)Find the absolute maximum value and the absolute minimum value of the function in the given interval:
iv) $f\left(x\right)=(x-1{)}^2+3,xϵ[-3,1]$

Answer 1

i) Given: $f\left(x\right)=x^3,xϵ[-2,2]$

Differentiation w.r.t $x$

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

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

$3x^2=0$

$\Rightarrow x^2=0$

$\Rightarrow x=0$

So, $x=0$ is a critical point.

Checking the functional values at $x=0$ and end points $x=-2,2$

We know, $f\left(x\right)=x^3$

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

$f(-2)=-8$

$f\left(2\right)=8$

Hence, the absolute maximum value of $f\left(x\right)$ is $8$ and the absolute minimum value of $f\left(x\right)$ is $-8$

ii) Given: $f\left(x\right)=\sin x+\cos x,xϵ[0,\pi ]$

Differentiation w.r.t $x$

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

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

$\Rightarrow \cos x-\sin x=0$

$\Rightarrow \cos x=\sin x$

$\Rightarrow \cot x=1$

$\Rightarrow x=\frac{\pi }{4}[\because xϵ[0,\pi \left]\right]$

So,

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

is a critical point.

Checking the functional values at

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

and end points $x=0,\pi$

We know, $f\left(x\right)=\sin x+\cos x$

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

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

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

$f\left(0\right)=\sin 0+\cos 0=1$

$f\left(\pi \right)=\sin \pi +\cos \pi =-1$

Hence, the absolute maximum value of $f\left(x\right)$ is $\sqrt{2}$ and the absolute minimum value of $f\left(x\right)$ is $-1$

iii) Given:

$f\left(x\right)=4x-\frac{1}{2}{x}^2$

Differentiation w.r.t $x$

$f^{\prime }\left(x\right)=4-\frac{1}{2}\times 2x$

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

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

$4-x=0$

$\Rightarrow x=4$

So, $x=4$ is a critical point.
Checking the functional values at $x=4$ and end points

$x=-2,\frac{9}{2}$

We know,

$f\left(x\right)=4x-\frac{1}{2}{x}^2$

$f\left(4\right)=4\left(4\right)-\frac{1}{2}(4{)}^2$

$\Rightarrow f\left(4\right)=16-\frac{1}{2}\times 16$

$\Rightarrow f\left(4\right)=16-8=8$

$f(-2)=4(-2)-\frac{1}{2}(-2{)}^2$

$\Rightarrow f(-2)=-8-\frac{1}{2}\times 4$

$\Rightarrow f(-2)=-8-2=-10$

$f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}{\left(\frac{9}{2}\right)}^2$

$\Rightarrow f\left(4\right)=18-\frac{1}{2}\times \frac{81}{4}$

$\Rightarrow f\left(4\right)=18-\frac{81}{8}=\frac{63}{8}$

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

is $8$ and the absolute minimum value of $f\left(x\right)$ is $-10$

iv) $f\left(x\right)=(x-1{)}^2+3$

Differentiation w.r.t $x$

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

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

$\Rightarrow 2(x-1)=0$

$\Rightarrow x=1$

So, $x=1$ is a critical point.
Checking the functional values at $x=1$ and end points $x=-3,1$

We know, $f\left(x\right)=(x-1{)}^2+3$

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

$f(-3)=(-3-1{)}^2+3$

$\Rightarrow f(-3)=16+3=19$

Hence, the absolute maximum value of $f\left(x\right)$ is $19$ and the absolute minimum value of $f\left(x\right)$ is $3$.