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

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

Answers (1)

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Answer:

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) <0then 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.

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