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  • Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17. 2 Question 2

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

Answers (1)

Answer:

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

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