The minimum value

Solution:

The given function is $f(x)=x^{2}+\frac{250}{x}, x \neq 0$.

$f(x)=x^{2}+\frac{250}{x}$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=2 x-\frac{250}{x^{2}}$

For maxima or minima,

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

$\Rightarrow 2 x-\frac{250}{x^{2}}=0$

$\Rightarrow x^{3}=\frac{250}{2}=125$

 

$\Rightarrow x=5$

Now,

$f^{\prime \prime}(x)=2+\frac{500}{x^{3}}$

At x = 5, we have

$f^{\prime \prime}(5)=2+\frac{500}{(5)^{3}}=2+\frac{500}{125}=2+4=6>0$

So, x = 5 is the point of local minimum.

$\therefore$ Minimum value of $f(x)=f(5)=(5)^{2}+\frac{250}{5}=25+50=75$        $\left[f(x)=x^{2}+\frac{250}{x}\right]$

Thus, the minimum value of the given function is 75.

The minimum value of $f(x)=x^{2}+\frac{250}{x}$ is ___75____.