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.
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