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  • Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 14

Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 14

Answers (1)

Answer:

x=2 is the point of local minima and the value of local minima is 2

Hint:

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

Given:

f(x)=\frac{x}{2}+\frac{2}{x}, x>0

Solution:-

f(x)=\frac{x}{2}+\frac{2}{x}

Differentiating f(x) w \cdot r . t^{\prime} x^{\prime}then

\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}\left\{\frac{2}{x}+\frac{x}{2}\right\} \\ &=\frac{d}{d x}\left\{\frac{2}{x}\right\}+\frac{d}{d x}\left\{\frac{x}{2}\right\} \quad\left[\frac{d}{d x}\{f(x)+h(x)\}=\frac{d}{d x}\{f(x)\}+\frac{d}{d x}\{h(x)\}\right] \\ &=2 \frac{d}{d x}\left(x^{-1}\right)+\frac{1}{2} \frac{d}{d x}(x) \quad\left[\frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ &=2(-1) x^{-1-1}+\frac{1}{2} x^{1-1} \quad\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \\ \end{aligned}

                       =-2 x^{-2}+\frac{1}{2} x^{0} \\

               =\frac{-2}{x^{2}}+\frac{1}{2} \\ f^{\prime}(x) =\frac{1}{2}-\frac{2}{x^{2}}

By first derivative test, for local maxima and local minima, we ha

f^{\prime}(x)=0

   \Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0 \Rightarrow \frac{1}{2}=\frac{2}{x^{2}}

\Rightarrow \quad x^{2}=4 \quad \Rightarrow x=\pm 2

\Rightarrow x=2             [x=-2$ is not possible since $x>0]

                                            -                      + 

                                   -∞              2               +∞       

Since {f}'\left ( x \right ) changes from -v e$ to $+v e when x increases through 2. So x=2 is the point of local minima

The value of local minima of f\left ( x \right ) at x=2  is

\begin{aligned} f(2) &=\frac{2}{2}+\frac{2}{2}=1+1 \\ &=2 \end{aligned}

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