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#### Need solution for RD Sharma maths class 12 chapter 17  Maxima and Minima exercise 17.3 question 5

x = 0 is local minima and its minimum value is 2 and x =1 is  local maxima and its maximum value is -6

Given:

$f(x)=\frac{4}{x+2}+x$

Explanation:

Differentiating f with respect to x

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(4(\mathrm{x}+2)^{-1}+\mathrm{x}\right) \\ &=-4(\mathrm{x}+2)^{-2}+1 \\ &=-\frac{4}{(x+2)^{2}}+1 \\ &=\frac{-4+\mathrm{x}^{2}+4 \mathrm{x}+4}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}(\mathrm{x}+4)}{(\mathrm{x}+2)^{2}} \end{aligned}

\begin{aligned} &\text { Put } f(x)=0 \\ &\frac{x(x+4)}{(x+2)^{2}}=0 \\ &x(x+4)=0 \\ &\text { i.e } x=0, \text { or } \\ &x=-4 \end{aligned}

Thus x = 0 and x = -4 are the points of local maxima and minima

Differentiating f’ with respect to x

\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}}\right) \\ &=\frac{(\mathrm{x}+2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+4 \mathrm{x}\right)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+2)^{2}}{(\mathrm{x}+2)^{4}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{(\mathrm{x}+2)^{2}(2 \mathrm{x}+4)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \cdot 2(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left[(\mathrm{x}+2)^{2}-\left(\mathrm{x}^{2}+4 \mathrm{x}\right)\right]}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left(\mathrm{x}^{2}+4 \mathrm{x}+4-\mathrm{x}^{2}-4 \mathrm{x}\right)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8}{(\mathrm{x}+2)^{3}} \end{aligned}

Put  x = 0 in f ‘’ (x)

\begin{aligned} \mathrm{f}^{\prime \prime}(0) &=\frac{8}{(0+2)^{3}} \\ &=\frac{8}{8} \\ &=1>0 \end{aligned}

S,x=0is local minima. Hence,its minimum value will be f(0)=2

And x= -4 in f ‘’(x)

\begin{aligned} &f^{\prime \prime}(-4)=\frac{8}{(-4+2)^{3}} \\ &=\frac{8}{-8} \\ &=-1<0 \end{aligned}

So x=-4 is local maxima. Hence, its maximum value will be f(-4)=-6