Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 4

X= 0 and x= -2 is the point of local minima and local maxima respectively.

The value of the local maxima and local minima is 0 and -4 respectively.

Hint:

Use first derivative test to find the point and value of the local maxima and local minima.

Given:

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

Solution:

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

Differentiating f(x) with respect to ‘x’ then

\begin{aligned} &\frac{d}{d x}\{f(x)\}=\frac{d}{d x}\left\{(x-1)(x+2)^{2}\right\} \\ &=(x-1) \frac{d}{d x}(x+2)^{2}+(x+2)^{2} \frac{d}{d x}(x-1) \\ &=(x-1) \cdot 2(x+2)+(x+2)^{2} \\ &=(x+2)(2 x-2+x+2) \\ &=(x+2)(3 x) \end{aligned}

By first derivative test,for local maxima or local minima ,we have

\begin{aligned} &\frac{d}{d x}\{f(x)\}=\frac{d}{d x}\left\{(x-1)(x+2)^{2}\right\} \\ &=(x-1) \frac{d}{d x}(x+2)^{2}+(x+2)^{2} \frac{d}{d x}(x-1) \\ &=(x-1) \cdot 2(x+2)+(x+2)^{2} \\ &=(x+2)(2 x-2+x+2) \\ &=(x+2)(3 x) \end{aligned}

+                          -                              +

-∞                    -2                      0                           ∞

Since, ${f}'\left ( x \right )$ changes from –ve to +ve when $x$ increases through 0. So $x=0$ is the point of local minima.

The value of the local minima of $f\left ( x \right )$ at $x=0$is

Again, Since ${f}'\left ( x \right )$ changes from +ve to -ve when $x$ increases through -2.

So, $x=-2$ is the point of local maxima.

The value of local maxima of $f\left ( x \right )$ at $x=-2$ is

$f\left ( -2 \right )=\left ( -2-1 \right )\left ( -2+2 \right )^{2}=-3\times 0=0$