How to solve this rational function 
for what real values of $a$ does the range of $f(x)=\frac{x+1}{a+x^2}$ contain the interval $[0,1]$?

I am anxious whether there can be a solution computed with derivatives something like this. I know SE demands is to show some effort and I got the answer but using very lengthy procedure and now I am anxious to know the solution using derivatives 
 A: Back to the pre-calculus theory of discriminants. Assuming $a\neq -1$ from now on (we have to deal with that case as a separate case), the equation
$$\frac{x+1}{a+x^2} = k \tag{1}$$
is equivalent to
$$ k x^2 - x+ (ka-1) = 0 \tag{2}$$
that has some real solution if its discriminant is non-negative, i.e. if
$$ 1-4k(ka-1) = 1+4k-4a k^2 \geq 0. $$
That gives that you may easily find the range of $f_a(x)=\frac{x+1}{a+x^2}$ without using derivatives.
Moreover, $1$ belongs to the range of $f_a(x)$ if $(1)$ or $(2)$ has some solution for $k=1$, i.e. iff:
$$ 5-4a\geq 0 \tag{3} $$
or $\color{red}{a\leq\frac{5}{4}}$. $0$ always belongs to the range of $f_a(x)$ since it is $f_a(-1)$.
I think you can easily finish from here: continuous functions over compact intervals have the intermediate value property, hence we just have to be careful with the case $a<0$ in which $f_a(x)$ has some real singularity. Anyway, the values of $k$ for which $1+4k-4ak^2$ vanishes are exactly the values of $f_a(x)$ in its stationary points, namely $\frac{1\pm\sqrt{1+a}}{2a}$. And we also have another interesting trick: $b$ belongs to the range of $f_a$ iff $\frac{1}{b}$ belongs to the range of $\frac{a+x^2}{x+1}$, that is the same as the range of $\frac{a+(x-1)^2}{x} = x+\frac{a+1}{x}-2$. That is way easier to study through the AM-GM inequality.
A: The method you cite will generally work.  I've taken some large steps in the following, but I bet you can fill in the gaps.  (The most commonly used tool is that "something squared" is nonnegative.)
For $a>0$, we compute $f'(x) = \dfrac{a-x(x+2)}{(x^2+a)^2}$.  This will have two zeroes at $-1 \pm \sqrt{a+1}$.  (We use the fact $a>0$ to notice that the denominator is always positive, so we do not have to worry about division by zero and $f$ having infinite limits.  In fact, we need only find the zeroes of the numerator, which is what we did.)  We plug these into $f$ and find the left one is always negative, $\dfrac{-\sqrt{a+1}}{1+(-\sqrt{a+1}-1)^2}$ (because the denominator is always positive) and the other is positive.  SO now we know $0$ is always in the range of $f$.  Checking the second derivative at the right stationary point, we get $f''(-1+\sqrt{a+1}) = \dfrac{(a+1)-1+\sqrt{a+1}}{2(-a-1+\sqrt{a+1})^3} = u$, which has a positive numerator and negative denominator, so the right stationary point is a local maximum.  As long as $u \geq 1$, the interval $[0,1]$ is in the range of $f$.  This happens for $a \leq 5/4$.  So, for all $a \in (0,5/4]$, the desired interval is in the range of $f$.
For $a=0$, the situation is much simpler.  $f(1) = \dfrac{2}{1} = 2$ and $\lim_{x \rightarrow \infty} f(x) = 0$ (although this value is never attained), so at least $(0,1]$ is in the range of $f$.  However, $f(-1) = 0$, so $[0,1]$ is in the range of this $f$.
For $a<0$, $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \sqrt{a}^+} = \infty$, and $f$ is continuous on $(\sqrt{a}, \infty)$, so we need only determine whether zero is in the range of $f$.  For $a \neq -1$, $f(-1) = \dfrac{0}{a+1} = 0$.
So all we need to know is the range of $\dfrac{x+1}{x^2-1} = \dfrac{1}{x-1}$.  Unfortunately, the range is $(-\infty, 0) \cup (0,\infty)$, so does not contain $[0,1]$.
Therefore, $[0,1]$ is in the range of $f$ for all $a \in (-\infty, -1) \cup (-1,5/4]$.
