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  • Need Solution for RD Sharma Maths Class 12 Chapter Mahima and Maxima Exercise Revision Exercise 17 point 2 Question sub question 1

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

Answers (1)

Answer:

  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                      

 

 

 

 

 

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