#### Need Solution for RD Sharma Maths Class 12 Chapter Mahima and Maxima Exercise Revision Exercise 17.2 Question Sub question 1

$x=5$ is the point of local minima and the value of local minimum of $f\left ( x \right )$ is 0.

Hint:

Use first derivative test to find the point and values of local maximum or local minimum.

Given:

$f\left ( x \right )=\left ( x-5 \right )^{4}$

Solution:

$f\left ( x \right )=\left ( x-5 \right )^{4}$

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

\begin{aligned} \frac{d}{d x}(f(x)) &=\frac{d}{d x}(x-5)^{4} \\ &=4(x-5)^{4-1} \frac{d}{d x}(x-5) \\ \end{aligned}

$=4(x-5)^{3}\left(\frac{d x}{d x}-\frac{d 5}{d x}\right) \\$                                  $\left[\therefore \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \quad \& \frac{d}{d x} \operatorname{constan} t=0\right]$

$=4(x-5)^{3}(1-0) \\$

$=4(x-5)^{3} \\$

$\therefore f^{\prime}(x)=+4(x-5)^{3}$

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

\begin{aligned} &\Rightarrow 4(x-5)^{3}=0 \\ &\Rightarrow \quad(x-5)^{3}=0 \\ &\Rightarrow \quad(x-5)=0 \\ &\Rightarrow \quad x=5 \end{aligned}

-∞                    -                    5                 +                  +∞

Since ${f}'\left ( x \right )$ changes from  -ve to +ve when  $x$ increases through 5.

so, $x$ = 5 is the point of local minima.

The value of local minima of $f\left ( x \right )$ at x = 5 is

$f\left ( 5 \right )=\left ( 5-5 \right )^{4}=0$