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#### Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 question 2

X=1 and x =-1 is the point of local minima and local maxima respectively and -2 and 2  is the value of local minima and local maxima respectively.

Hint:

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

Given:

$f\left ( x \right )=x^{3}-3x$

Solution:

$f\left ( x \right )=x^{3}-3x$

Differentiating $f\left ( x \right )$ with respect to ‘x’ then

$\frac{d(f(x))}{d x}=\frac{d}{d x}\left(x^{3}-3 x\right)=\frac{d\left(x^{3}\right)}{d x}-\frac{d(3 x)}{d x} \quad$                                     $\left[\because \frac{d}{d x}(a x+b)=\frac{d(a x)}{d x}+\frac{d(b)}{d x}\right]$

$=\frac{d\left(x^{3}\right)}{d x}-3 \frac{d(x)}{d x} \quad$                       $\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d\left(x^{n}\right)}{d x}\right]$

$=3 x^{3-1}-3 x^{1-1} \quad$                          $\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$

$=3x^{2}-3x^{0}$

$=3x^{2}-3$                                       $\left [ \because x^{0}=1 \right ]$

$\because {f}'\left ( x \right )=+3\left ( x^{2}-1 \right )$

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

$\because {f}'\left ( x \right )=0$

$\begin{array}{ll} \Rightarrow 3\left(x^{2}-1\right)=0 & \Rightarrow\left(x^{2}-1\right)=0 \quad[\because 3 \neq 0] \\ \Rightarrow x^{2}=1 & \Rightarrow x=\pm 1 \end{array}$

−                       +

-∞                          -1                                 +1                        ∞

Since   ${f}'\left ( x \right )$ changes from –ve  to +ve  as $x$ increases through 1.

So, $x$ = 1 is the point of local minima.

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

$f\left ( 1\right )=1^{3}-3\cdot 1=3-1=-2$

Again, since ${f}'\left ( x \right )$ changes from +ve to  -ve as $x$ 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

$f\left (-1\right )=\left ( -1 \right )^{3}-3\left ( -1 \right )$

$-1+3=2$