Practical and traditional way to get the maximum of $\frac{x^2}{(x^4+1)}$? The maximum seems to be $\dfrac{1}{2}$, but how do you get this value and why? Does it have anything to do with the graph of the function?
 A: By the AM-GM inequality
$$X^2 \leq \frac{X^4+1}{2} \Rightarrow  \frac{X^2}{X^4+1} \leq \frac{1}{2}$$
A: Hint: This is a textbook example of techniques from Calculus I, but if you don't know calculus this method probably isn't useful to you.
Since this function is differentiable everywhere on the real line, just find where the derivative vanishes. These are the only candidates for local extrema. Compute the value at each of these and pick the largest. No need to bother with 2nd derivative because the list is so short, evaluating the function is easier than computing the second derivative. You should also note the the function tends to zero toward $\pm\infty$, so there is a maximum because the function is nonnegative.
A: you can do this with precalculus. look at the $$y = \frac{x^2}{1 + x^4}  .$$ turn this into quadratic equation $$yu^2 - u + y= 0 $$ for $u = x^2.$ for this quadratic equation to have real solutions you need the discriminant $$1 - 4y^2 \ge 0 \text{ and } y \ge 0 .$$  that gives you the maximum $y$ to be $\dfrac{1}{2}.$
A: We have $\displaystyle f(x)=\frac{x^2}{1+x^4}$, with $\displaystyle f'(x)=\frac{2x(x^4-1)}{(1+x^4)^2}$ and $\displaystyle f''(x)=\frac{2-24x^4+6x^8}{(1+x^4)^3}$. $\displaystyle f'(x)=0$ results in $x=0,-1,1$, where $\displaystyle f''(0)=2>0, f''(1)=-2<0, f''(-1)=-2<0$. Therefore $\displaystyle f(x)$ is maximized at $x=1,-1$ with the maximum value of $\displaystyle f(1)=f(-1)=\frac12 $.
Note that for $x=\infty$ and $x=-\infty$ you also get $f'(x)=0$ hence these are also extremums. But these are neither maximums nor minimums since $f''(\infty)=0$ and $f''(-\infty)=0$.
