Find the maximum and minimum values of the function f(x)=4 x+2 +x - Vidyadip

Question

Find the maximum and minimum values of the function $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$

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Answer

Given, $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$
$\Rightarrow\text{f}'(\text{x})=-\frac{4}{(\text{x}+2)^{2}}+1$
For a local maxima or a local minima, We must have f'(x) = 0
$\Rightarrow-\frac{4}{(\text{x}+2)^{2}}+1=0$
$\Rightarrow-\frac{4}{(\text{x}+2)^{2}}=-1$
$(\text{x}+2)^{2}=\pm2$
$\Rightarrow \text{x}=0 \ \text{and} -4$
Thus, x = 0 and x = -4 are the possible of local maxima or local minima.
Now, $\text{f}'(\text{x})=\frac{8}{(\text{x}+2)^{3}}$
At x = 0
$\text{f}''(0)=\frac{8}{(2)^{3}}=1>0$
So, x = 0 is a point of local minimum.
The local minimum value is given by
$\text{f}(0)=\frac{4}{(0+2)}+0=2$

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