Question

Determine the minimum value of the function:
$f\left(x\right)=x^2+\frac{16}{x^2}$

Answer 1

$f\left(x\right)=x^2+\frac{16}{x^2}{f}^{\prime }\left(x\right)=2x-\frac{32}{x^3}\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow 2x-\frac{32}{x^3}=02x^4=32\Rightarrow 2^4=16\Rightarrow x\pm 2f\left(x\right)={\left(2\right)}^2+\frac{16}{2^2}=4+4=8f^{\prime \prime }\left(x\right)=2-\frac{96}{x^4}>0$

Therefore, $f\left(x\right)$ will have minimum

So, The minimum value of $f\left(x\right)=8$