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

Point of local maxima is at x=1 and it’s local maximum value is 68. Also point of local minima is at x=5,-6 and it’s local minimum values are -316 and -1647.

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:

$f(x)=x^{4}-62 x^{2}+120 x+9$

Explanation:

It is given that

\begin{aligned} &f(x)=x^{4}-62 x^{2}+120 x+9 \\ &f^{\prime}(x)=4 x^{3}-62 * 2 x+120 \end{aligned}

To find critical values ,

\begin{aligned} &f^{\prime}(x)=0 \\ &4 x^{3}-124 x+120=0 \\ &4\left(x^{3}-31 x+30\right)=0 \end{aligned}

On solving,

critical values of x are 5,1,-6.

Now,

$f^{\prime \prime}(x)=12 x^{2}-1$

At x=5,

\begin{aligned} f^{\prime \prime}(5) &=12(5)^{2}-124 \\ &=176>0 \end{aligned}

So, x=5 is point of local minima.

At x=1,

\begin{aligned} f^{\prime \prime}(1) &=12(1)^{2}-124 \\ &=-112<0 \end{aligned}

So, x=1 is point of local maxima.

At x=-6,

\begin{aligned} f^{\prime \prime}(-6) &=12(-6)^{2}-124 \\ &=308>0 \end{aligned}

So,  x=-6 is a point of local minima.

The local max. value at x=1 is

\begin{aligned} f(1) &=(1)^{4}-62(1)^{2}+120(1)+9 \\ &=68 \end{aligned}

And local min. value at x=5 & x=-6 are,

\begin{aligned} f(5)=&(5)^{4}-62(5)^{2}+120(5)+9 \\ &=-316 \\ f(-6) &=(-6)^{4}-62(-6)^{2}+120(-6)+9 \\ &=-1647 \end{aligned}

Hence, point of local maxima is 1 & its maximum value is 68.

Also, point of local minima are 5&-6 and its minimum values are -316 &-1647 respectively.