#### Please solve R.D.Sharma class 12 Chapter 17 Maxima and Minima excercise 17.3 question 1 Sub Question 3 maths textbook solution.

Point of local maxima value is -2 and it’s local maximum value is 0. Also point of local minima is 0and it’s local minimum value is  -4

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If $f''(c_1) >0$ then $c_1$ is point of local minima.

If $f''(c_2) <0$then $c_2$ is point of local maxima .

where $c_1$ &$c_2$ are critical points.

Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.

Given:

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

Explanation:

We have,

\begin{aligned} f(x) &=(x-1)(x+2)^{2} \\ f^{\prime}(x) &=(x+2)^{2}(1)+(x-1) 2(x+2) \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=(x+2)(x+2+2 x-2) \\ &=(x+2)(3 x) \\ f^{\prime \prime}(x) &=3 x(1)+(x+2) 3 \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=6 x+6 \end{aligned}

For max and min, f’(x)=0,

\begin{aligned} &(x+2)(3 x)=0 \\ &x=0 \& x=-2 \end{aligned}

At x=0,

$f^{\prime \prime}(0)=6(0)+6=6>0$

So, x= 0 is point of local minima

$f^{\prime \prime}(0)=6(0)+6=6>0$

At x=-2,

\begin{aligned} f^{\prime \prime}(0) &=6(-2)+6 \\ &=-6<0 \end{aligned}

So, x = -2 is point of local maxima

So, local max. value at x=-2 is

f(-2)=(-2-1)(-2+2)^{2}=0

And local min. value at x= 0 is

$f(0)=(0-1)(0+2)^{2}=-4$

Thus, point of local maxima is -2 & its max. value is 0 & point of local minima is 0 & its value is -4.