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Question:

If $(x)=x+\frac{1}{x}, x>0$, then its greatest value is

(a) $-2$

(b) 0

(c) 3

(d) none of these

Solution:

Given : $f(x)=x+\frac{1}{x}$

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

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

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

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

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$

$\Rightarrow x=1$              (Given : $x>0$ )

Now,

$f^{\prime \prime}(x)=\frac{2}{x^{3}}$

$\Rightarrow f^{\prime \prime}(1)=2>0$

So, $x=1$ is a local minima.