Application of Derivatives — Class 12 Maths Solution

(i) We have, $f(x) = {x^2} \Rightarrow f'(x) = 2x$
For critical points,
$f'(x) = 0 \Rightarrow 2x = 0 \Rightarrow x = 0$

The point at which extremum may occur is $x = 0$
Now, $f''(x) = 2 \Rightarrow f''(0) = 2 > 0$
Therefore, $f$ has a local minima at $x = 0$ and local minimum value is $f(0) = {0^2} = 0$.

(ii) We have, $g(x) = {x^3} - 3x \Rightarrow g'(x) = 3{x^2} - 3$

For critical points$,g(x) = 0$
$\Rightarrow 3{x^2} - 3 = 0 \Rightarrow {x^2} = 1 \Rightarrow x = 1, - 1$
The points at which extremum may occur are $- 1$ and $1$ .
$g''(x) = 6x$

$\Rightarrow g''( - 1) = 6( - 1) = - 6 < 0$
Therefore, $g$ has a local maxima at $x = - 1$ and local maximum value at $x = - 1$ is $g( - 1) = {( - 1)^3} - 3( - 1) = - 1 + 3 = 2$.

$g''(1) = 6 \times 1 = 6 > 0$
Therefore, $g$ has a local minima at $x = 1$ and local minimum value at $x = 1$ is $g(1) = {1^3} - 3 \times 1 = - 2$.

(iii) We have, $h(x) = \sin x + \cos x,0 < x < \cfrac{\pi }{2}$
$\Rightarrow h'(x) = \cos x - \sin x$ for all $x \in \left( {0,\;\cfrac{\pi }{2}} \right)$

For critical points, $h'(x) = 0$
$\Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \cfrac{\pi }{4}$

The point at which extremum may occur is $x = \cfrac{\pi }{4}$
$h''(x) = - \sin x - \cos x$

$\Rightarrow h''\left( {\cfrac{\pi }{4}} \right) = - \sin \cfrac{\pi }{4} - \cos \cfrac{\pi }{4} = \cfrac{{ - 2}}{{\sqrt 2 }} < 0$

Therefore, $h$ has a local maxima at $x = \cfrac{\pi }{4}$ and local maximum value at $x = \cfrac{\pi }{4}$ is $h\left( {\cfrac{\pi }{4}} \right) = \sin \cfrac{\pi }{4} + \cos \cfrac{\pi }{4} = \cfrac{1}{{\sqrt 2 }} + \cfrac{1}{{\sqrt 2 }} = \cfrac{2}{{\sqrt 2 }} = \sqrt 2$.

(iv) We have,
$f(x) = \sin x - \cos x,\;0 < x < 2\pi$
$\Rightarrow f'(x) = \cos x + \sin x$

For critical points, $f'(x) = 0$
$\Rightarrow \cos x + \sin x = 0 \Rightarrow \tan x = - 1$

$\Rightarrow x = \pi - \cfrac{\pi }{4},2\pi - \cfrac{\pi }{4} \Rightarrow x = \cfrac{{3\pi }}{4},\cfrac{{7\pi }}{4}$

Therefore, the points at which extremum may occur are $x$
$x = \cfrac{{3\pi }}{4}$ and $x = \cfrac{{7\pi }}{4}$
$f''(x) = - \sin x + \cos x$

$\Rightarrow f''\left( {\cfrac{{3\pi }}{4}} \right) = - \sin \cfrac{{3\pi }}{4} + \cos \cfrac{{3\pi }}{4} = - \cfrac{1}{{\sqrt 2 }} - \cfrac{1}{{\sqrt 2 }} < 0$

Therefore,$f$ has local maxima at $x = \cfrac{{3\pi }}{4}$ and local maximum value at $x = \cfrac{{3\pi }}{4}$ is $f\left( {\cfrac{{3\pi }}{4}} \right) = \sin \cfrac{{3\pi }}{4} - \cos \cfrac{{3\pi }}{4} = \cfrac{1}{{\sqrt 2 }} - \left( { - \cfrac{1}{{\sqrt 2 }}} \right) = \cfrac{2}{{\sqrt 2 }} = \sqrt 2$.

Further $f''\left( {\cfrac{{7\pi }}{4}} \right) = - \sin \cfrac{{7\pi }}{4} + \cos \cfrac{{7\pi }}{4}$

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

Therefore, $f$ has local minima at $x = \cfrac{{7\pi }}{4}$ and local minimum value at $x = \cfrac{{7\pi }}{4}$ is $f\left( {\cfrac{{7\pi }}{4}} \right) = \sin \cfrac{{7\pi }}{4} - \cos \cfrac{{7\pi }}{4} = \cfrac{{ - 1}}{{\sqrt 2 }} - \cfrac{1}{{\sqrt 2 }} = - \sqrt 2$.

(v) We have, $f(x) = {x^3} - 6{x^2} + 9x + 15,x \in R$
$\Rightarrow f'(x) = 3{x^2} - 12x + 9,x \in R$
For critical points, $f'(x) = 0$
$\Rightarrow 3{x^2} - 12x + 9 = 0 \Rightarrow 3(x - 1)(x - 3) = 0 \Rightarrow x = 1,x = 3$

Therefore, the points at which extremum may occur are $x = 1$ and $x = 3$.
$f''(x) = 6x - 12,\,\,x \in R. \Rightarrow f''(1) = 6 \times 1 - 12 = - 6 < 0$

Therefore we can say that $f$ has a local maxima at $x = 1$ and local maximum value at $x = 1$ is
$f(1) = 1 - 6 + 9 + 15 = 19$.
$f''(3) = 6 \times 3 - 12 = 6 > 0$

Therefore, $f$ has a local minima at $x = 3$ and local minimum value at $x = 3$ is $f(3) = {3^3} - 6 \times {3^2} + 9 \times 3 + 15 = 15$.

(vi) Given, $g(x) = \cfrac{x}{2} + \cfrac{2}{x},x > 0 \Rightarrow g'(x) = \cfrac{1}{2} + \left( { - \cfrac{2}{{{x^2}}}} \right),x > 0$

For critical points, $g'(x) = 0$
$\Rightarrow \cfrac{1}{2} - \cfrac{2}{{{x^2}}} = 0 \Rightarrow {x^2} - 4 = 0 \Rightarrow x = - 2,2$

Therefore the only point where extremum may occur is $x = 2.$
$g''(x) = - 2( - 2){x^{ - 3}},x > 0$ and $g''(2) = 4{(2)^{ - 3}} = \cfrac{4}{8} = \cfrac{1}{2} > 0$

Therefore we can say that $f$ has a local minima at $x = 2$ and local minimum value is $g(2) = \cfrac{2}{2} + \cfrac{2}{2} = 2$

(vii) Given,
$\Rightarrow g'(x) = \cfrac{{ - 2x}}{{{{\left( {{x^2} + 2} \right)}^2}}}$
${({x^2} + 2)^2}$For critical points, $g'(x) = 0$

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

$g''(x) = \cfrac{{6{x^2} - 4}}{{{{({x^2} + 2)}^3}}} \Rightarrow g''(0) = \cfrac{{ - 4}}{8} < 0$

Therefore we can say that $g$ has a local maxima at $x = 0$ and local maximum value is $g(0) = \cfrac{1}{{0 + 2}} = \cfrac{1}{2}$

(viii) We have, $f(x) = x\sqrt {1 - x} ,0 < x \le 1$

$\Rightarrow f'(x) = \cfrac{{x( - 1)}}{{2\sqrt {1 - x} }} + \sqrt {1 - x} = \cfrac{{ - x + 2(1 - x)}}{{2\sqrt {1 - x} }} = \cfrac{{2 - 3x}}{{2\sqrt {1 - x} }}$

For critical points, $f'(x) = 0 \Rightarrow \cfrac{{2 - 3x}}{{2\sqrt {1 - x} }} = 0 \Rightarrow x = \cfrac{2}{3}$.

Therefore we can say that the point at which extremum may occur is $x = 2/3$.
$f''(x) = \cfrac{{\cfrac{1}{2}\left\{ {(\sqrt {1 - x} )( - 3) - \cfrac{{(2 - 3x)( - 1)}}{{2\sqrt {1 - x} }}} \right\}}}{{(1 - x)}}$

$= \cfrac{1}{2}\left[ {\cfrac{{2(1 - x)( - 3) + 2 - 3x}}{{2\sqrt {1 - x} (1 - x)}}} \right]$

$\Rightarrow f''\left( {\cfrac{2}{3}} \right) = \cfrac{1}{2}\left[ {\cfrac{{2\left( {1 - \cfrac{2}{3}} \right)( - 3) + 2 - 3 \times \left( {\cfrac{2}{3}} \right)}}{{2\sqrt {1 - \cfrac{2}{3}} \left( {1 - \cfrac{2}{3}} \right)}}} \right]$

$= \cfrac{1}{2}\left( {\cfrac{{\cfrac{{ - 2 \times 1 \times 3}}{3} + 2 - 2}}{{2\sqrt {\cfrac{1}{3}} \left( {\cfrac{1}{3}} \right)}}} \right) < 0$

Therefore, $f$ has local maxima at $x = \cfrac{2}{3}$ and local maximum value is $f\left( {\cfrac{2}{3}} \right) = \cfrac{2}{3}\sqrt {\cfrac{1}{3}} = \cfrac{2}{{3\sqrt 3 }} = \cfrac{{2\sqrt 3 }}{9}$