Get Answers to all your Questions

header-bg qa

Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 6

Answers (1)


x= 3 and x= 1 is the point of  local minima and  local maxima respectively. The value of local  minima and maxima is 15 and 19 respectively.


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


f\left ( x \right )=x^{3}-6x^{2}+9x+15


f\left ( x \right )=x^{3}-6x^{2}+9x+15

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

\frac{d}{d x}\{f(x)\}=\frac{d}{d x}\left\{x^{3}-6 x^{2}+9 x+15\right\} \quad\left[\because \frac{d}{d x}\left(a x^{3}+b x^{2}+c x+d\right)=\frac{d}{d x}\left(a x^{3}\right)+\frac{d}{d x}\left(b x^{2}\right)+\frac{d}{d x}(c x)+\frac{d}{d x}(d)\right]

                      =\frac{d}{d x}\left(x^{3}\right)-\frac{d}{d x}\left(6 x^{2}\right)+\frac{d}{d x}(9 x)+\frac{d}{d x}(15)

                      =\frac{d}{d x}\left(x^{3}\right)-6 \frac{d}{d x}\left(x^{2}\right)+9 \frac{d}{d x}(x)+\frac{d}{d x}(15)\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right]

                      =3 x^{3-1}-6.2 x^{2-1}+9 x^{1-1}+0 \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}, \frac{d}{d x} \operatorname{constan} t=0\right]

\therefore f^{\prime}(x)=3 x^{2}-12 x+9 x=3\left(x^{2}-4 x+3\right)

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

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 3\left(x^{2}-4 x+3\right)=0 \Rightarrow\left(x^{2}-4 x+3\right)=0 \quad[\because 3 \neq 0] \\ &\Rightarrow(x-1)(x-3)=0 \quad \Rightarrow x-1=0 \quad \text { or } x-3=0 \\ &\Rightarrow x=1 \quad \text { or } \quad x=3 \end{aligned}    

                                                       +                -                    + 

                                   -∞                          1                    3                          ∞


 sincef^{\prime}(x) changes from -ve to +ve when x increases through 3.

So, x=3 is the point of local minima

The value of local minima of f\left ( x \right ) at x=3 is

\begin{aligned} f(3) &=(3)^{3}-6.3^{2}+9.3+15 \\ &=27-6.9+27+15 \\ &=54-54+15 \\ &=15 \end{aligned}

Again since f^{\prime}(x) changes from +ve to -ve when x increases through 1.

So, x=1 is the point of local maxima

The value of local maxima of f\left ( x \right ) at x=1 is

\begin{aligned} f(1) &=(1)^{3}-6.1^{2}+9.1+15 \\ &=1-6+9+15 \\ &=25-6 \\ &=19 \end{aligned}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support