1. Determine the minimum and maximum value of the function $f(x)=\frac{x^{2}}{a x^{4}+b}$, where $a, b>0$, and then determine for which values of the argument $x$ they are achieved.
Solution 1. Clearly $x^{2} \geq 0$, and from the condition $a, b>0$ it follows that $a x^{4}+b>0$. Therefore $f(x) \geq 0$, so the minimum value of the function is $f(x)=0$ and it is achieved only for $x=0$.

From the inequality between the arithmetic and geometric mean for the positive numbers $a x^{4}$ and $b$ we get $a x^{4}+b \geq 2 \sqrt{c x^{4} \cdot b}=2 x^{2} \sqrt{a b}$. Therefore, $\frac{x^{2}}{a x^{4}+b^{2}} \leq \frac{x^{2}}{2 x^{2} \sqrt{a b}}=\frac{1}{2 \sqrt{a b}}$. Thus, the maximum value of the given function is $\frac{1}{2 \sqrt{a b}}$. The maximum is achieved when the inequality $a x^{4}+b \geq 2 x^{2} \sqrt{a b}$ becomes an equality, i.e., when $a x^{-1}=b$. Solving the equation $c x^{4}=b$, we get $x= \pm \sqrt{\frac{b}{a}}$.