Write the minimum value of f(x)=x+1 x ,x>0. - Vidyadip

Question

Write the minimum value of $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}},\text{x}>0.$

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Answer

Given, $\text{f}\text{(x)}=\text{x}+\frac{1}{\text{x}}$
$\therefore\ \text{f}'\text{(x)}=1-\frac{1}{\text{x}^2}$
For a local maxima or a local minima,
$\text{f}'\text{(x)}=0$
$\Rightarrow1-\frac{1}{\text{x}^2}=0$
$\Rightarrow\text{x}= +1,-1$
Now, $\text{f}''(\text{x})=\frac{2}{\text{x}^{3}}$
At x = 1, f''(x)
x = -1, f''(x) = -2 < 0
$\therefore $ x = -1 is point of local minima
$\therefore $ maximum value of = f(-1) = -2.

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