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

$x= 3$ 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.

Hint:

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

Given:

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

Solution:

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

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                          ∞

since$f^{\prime}(x)$ 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}