Get Answers to all your Questions

Name: Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise 17.3 Question 5 Maths Textbook Solution.
Uploaded: Mon, 2021-09-27 16:13
Description: Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise 17.3 Question 5 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise 17.3 Question 5 Maths Textbook Solution.

Answers (1)

Answer:

x = 0 is local minima and its minimum value is 2 and x =-4 is local maxima and its maximum value is -6

Given:

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

Explanation:

Differentiating f with respect to x

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

$\begin{aligned} &=\frac{-4+x^{2}+4 x+4}{(x+2)^{2}} \\ &=\frac{x^{2}+4 x}{(x+2)^{2}} \\ &=\frac{x(x+4)}{(x+2)^{2}} \end{aligned}$

Put $f\left ( x \right )=0$

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

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

Differentiating f’ with respect to x

$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}}\right)$

$=\frac{(\mathrm{x}+2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+4 \mathrm{x}\right)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+2)^{2}}{(\mathrm{x}+2)^{4}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}}$