The function

The given function is $f(x)=\frac{x}{2}+\frac{2}{x}, x \neq 0$.

$f(x)=\frac{x}{2}+\frac{2}{x}$

Differentiating both sides with respect to x, we get

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

For maxima or minima,

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

At x = −2, we have

$f^{\prime \prime}(-2)=\frac{4}{(-2)^{3}}=-\frac{1}{2}<0$

So, $x=-2$ is the point of local maximum of $f(x)$.

At x = 2, we have

$f^{\prime \prime}(2)=\frac{4}{(2)^{3}}=\frac{1}{2}>0$

So, x = 2 is the point of local minimum of f(x).

Thus, the given function $f(x)=\frac{x}{2}+\frac{2}{x}$ has a local minimum at $x=2$

The function $f(x)=\frac{x}{2}+\frac{2}{x}$ has a local minimum at $x=$ ____2____.