Application of Derivatives — Class 12 Maths Solution

We have:$f(x) = {(x - 2)^4}{(x + 1)^3}$
$f'(x) = {(x - 2)^4} \cdot 3{(x + 1)^2} + {(x + 1)^3} \cdot 4{(x - 2)^3}$

$= {(x - 2)^3}{(x + 1)^2}[3(x - 2) + 4(x + 1)]$
$= {(x - 2)^3}{(x + 1)^2}(7x - 2) = 7{(x - 2)^3}{(x + 1)^2}\left( {x - \cfrac{2}{7}} \right)$

To find maximum or minimum, $f'(x) = 0$
$\Rightarrow 7{(x - 2)^3}{(x + 1)^2}\left( {x - \cfrac{2}{7}} \right) = 0 \Rightarrow x = 2, - 1,\cfrac{2}{7}$

(ii) When $x < 2$ (slightly), $f'(x) = ( - )( + )( + ) = - ve$

When $x > 2$ (slightly), $f'(x) = ( + )( + )( + ) = + ve$

Therefore, $f'(x)$ changes its sign from $- ve$ to $+ ve$ when passing through the point $x = 2$.

Thus, has local minima at $x = 2.$

(iii) When $x < - 1$ (slightly), $f'(x) = ( - )( + )( - ) = + ve$

When $x > - 1$ (slightly), $f'(x) = ( - )( + )( - ) = + ve$

Therefore, $f'(x)$ does not change its sign when passing through the point $x = - 1$. Thus, $x = - 1$ is a point of inflexion.

(i) When $x < \cfrac{2}{7}$ (slightly), $f'(x) = ( - )( + )( - ) = + ve$

When $x > \cfrac{2}{7}$ (slightly), $f'(x) = ( - )( + )( + ) = - ve$

Therefore, $f'(x)$ changes its sign from $+ ve$ to $- ve$ $\dot w$hen passing through the point $x = \cfrac{2}{7}$. Thus, $f(x)$ has local maxima at $x = \cfrac{2}{7}.$