Question:
$f(x)=\frac{x}{2}+\frac{2}{x}, x>0$
Solution:
Given: $f(x)=\frac{x}{2}+\frac{2}{x}$
$\Rightarrow f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$
For the local maxima or minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0$
$\Rightarrow \frac{1}{2}=\frac{2}{x^{2}}$
$\Rightarrow x^{2}=\pm 2$
Since $x>0, f^{\prime}(x)$ changes from negative to positive when $x$ increases through $2 .$ So, $x=2$ is a point of local minima.
The local minimum value of $f(x)$ at $x=2$ is given by
$\frac{2}{2}+\frac{2}{2}=2$
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