Question

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

Answer 1

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

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\frac{1}{2}-\frac{2}{x^2}$

For maxima or minima,

$f\text{'}\left(x\right)=0$

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

$\Rightarrow x^2=4$

⇒ x = −2 or x = 2

Now,

$f\text{'}\text{'}\left(x\right)=\frac{4}{x^3}$

At x = −2, we have

$f\text{'}\text{'}\left(-2\right)=\frac{4}{{\left(-2\right)}^3}=-\frac{1}{2}<0$

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

At x = 2, we have

$f\text{'}\text{'}\left(2\right)=\frac{4}{{\left(2\right)}^3}=\frac{1}{2}>0$

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____.