#### Please solve R.D.Sharma class 12 Chapter 17  Maxima and Minima excercise 17.3 question 1 sub question 2 maths textbook solution.

point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.

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:

$f(x)=x^{3}-6 x^{2}+9 x+15$

Explanation:

We have,

\begin{aligned} f(x) &=x^{3}-6 x^{2}+9 x+15 \\ f^{\prime}(x) &=3 x^{2}-6.2 x+9 \\ &=3 x^{2}-12 x+9 \\ \therefore f^{\prime}(x) &=3\left(x^{2}-4 x+3\right) \\ \& f^{\prime \prime}(x) &=3(2 x-4) \\ &=6 x-12 \end{aligned}

To find maxima and minima.

\begin{aligned} &\qquad f^{\prime}(x)=0 \\ &3\left(x^{2}-4 x+3\right)=0 \\ &\quad x^{2}-4 x+3=0 \\ &X=3 \text { or } x=1 \\ &\text { At } x=3, \\ &f^{\prime \prime}(3)=6(3)-12 \\ &=6>0 \end{aligned}

At x = 3,

f’’(3) = 6(3)-12

=6>0

So, x = 3 is point of local minima

At x = 1,

f’’(1) = 6(1)-12

=-6<0

So, x = 1 is point of local maxima

Now, local maximum value at x= 1 is

\begin{aligned} &f(1)=(1)^{3}-6(1)^{2}+9(1)+15 \\ &f(1)=19 \end{aligned}

And local min. value at x=3 is

\begin{aligned} &f(3)=(3)^{3}-6(3)^{2}+9(3)+15 \\ &f(3)=15 \end{aligned}

Thus, point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.