Application of Derivatives — Class 12 Maths Solution

(i) $f\left( x \right) = {x^3},x \in [ - 2,2] \& \Rightarrow \,\,f'(x) = 3{x^2}$

For critical points, $f'\left( x \right) = 0 \Rightarrow 3{x^2} = 0 \Rightarrow x = 0 \in [ - 2,2]$

Hence, for finding the absolute maximum value and the absolute minimum value, we have to evaluate $f(0),f( - 2)$ and $f(2)$.

$f\left( 0 \right) = {0^3} = 0,\,f\left( { - 2} \right) = {\left( { - 2} \right)^3} = - 8$ and $f\left( 2 \right) = {2^3} = 8$

Therefore the absolute maximum value of $f\left( x \right)$ is $8$ at $x = 2$and absolute minimum value of $f\left( x \right)$is $- 8$ at $x = - 2$.

(ii) We have, $f\left( x \right) = \sin x + \cos x,\,x \in [0,\pi ]$
$\Rightarrow \,\,f'(x) = \cos x - \sin x$
For critical points, $f'\left( x \right) = 0$
$\Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \cfrac{\pi }{4} \in [0,\pi ]$

Hence, for finding the absolute maximum value and the absolute minimum value, we have to evaluate $f(0),f(\pi )$ and $f\left( {\cfrac{\pi }{4}} \right)$.
$f\left( 0 \right) = \sin 0 + \cos 0 = 0 + 1 = 1,\,$

$f\left( \pi \right) = \sin \pi + \cos \pi = 0 - 1 = - 1$
$f\left( {\cfrac{\pi }{4}} \right) = \sin \cfrac{\pi }{4} + \cos \cfrac{\pi }{4} = \cfrac{1}{{\sqrt 2 }} + \cfrac{1}{{\sqrt 2 }} = \sqrt 2$

Therefore the absolute maximum value of $f\left( x \right)$ is $\sqrt 2$ at $x = \cfrac{\pi }{4}$and absolute minimum value of $f\left( x \right)$is $- 1$ at $x = \pi$

(iii) We have , $f\left( x \right) = 4x - \cfrac{1}{2}{x^2},x \in \left[ { - 2,\cfrac{9}{2}} \right]$
$\Rightarrow f'\left( x \right) = 4 - x$
For critical points, $f'\left( x \right) = 0 \& \Rightarrow 4 - x = 0$
$\Rightarrow x = 4 \in \left[ { - 2,\cfrac{9}{2}} \right]$

To find the absolute maximum value and the absolute minimum value, we have to evaluate $f( - 2),f\left( {\cfrac{9}{2}} \right)$ and $f\left( 4 \right)$.
$f\left( { - 2} \right) = 4\left( { - 2} \right) - \cfrac{1}{2}{\left( { - 2} \right)^2} = - 8 - 2 = - 10,$

$f\left( {\cfrac{9}{2}} \right) = 4\left( {\cfrac{9}{2}} \right) - \cfrac{1}{2}{\left( {\cfrac{9}{2}} \right)^2} = 18 - \cfrac{{81}}{8} = \cfrac{{63}}{8}$

$f\left( 4 \right) = 4 \times 4 - \cfrac{1}{2}{\left( 4 \right)^2} = 16 - 8 = 8$

Therefore the absolute maximum value of $f\left( x \right)$ is $8$ at $x = 4$and absolute minimum value of $f\left( x \right)$is $- 10$ at $x = - 2$

(iv) We have , $f\left( x \right) = {(x - 1)^2} + 3,x \in [ - 3,1]$
$\Rightarrow f'\left( x \right) = 2(x - 1)$
For critical points, $f'\left( x \right) = 0$
$\Rightarrow 2(x - 1) = 0 \Rightarrow x = 1 \in \left[ { - 3,1} \right]$

Hence, for finding the absolute maximum value and the absolute minimum value of $f\left( x \right)$ , we have to evaluate $f\left( { - 3} \right)$ and $f\left( 1 \right)$.

$f\left( { - 3} \right) = {\left( { - 3 - 1} \right)^2} + 3 = 19$and $f\left( 1 \right) = {\left( {1 - 1} \right)^2} + 3 = 3$

Therefore the absolute maximum value of $f\left( x \right)$ is $19$ at $x = - 3$and absolute minimum value of $f\left( x \right)$is $3$ at $x = 1$