Given functions $f(x)$ and $g(x)$ find all values of $a$ such that domain of $f(x)$ includes the range of $g(x)$
As in the title i'm given two functions:
$$ f(x) = \sqrt{\frac{2a+x}{a-x}} \\ g(x) = \frac{1}{x^2 + 2x +4a - 2} $$
The domain of $f(x)$ is $x \in [-2a, a), \; a>0 \;$ or $x \in (a, -2a], \; a < 0$. But how do i find the range of $g(x)$ given $a$ is unknown (for example when denominator has zeros, then the function tends to infinity)?
Basically i'm interested in a general approach. I guess one could somehow translate this into a system of inequalities and derive $a$ from there, but not sure.
I'm not supposed to use "calculus" for this problem, but would still appreciate any hints disregarding the chosen method.
How do I solve such kind of problems?
When $x^2 + 2x + 4a - 2$ has two distinct (real) roots, this is going to lead to two asymptotes. Basically, your function is going to be of the form $$\frac{1}{(x - x_1)(x - x_2)},$$ where $x_1 < x_2$, which will lead to two asymptotes at these two roots. When $x_1 < x < x_2$, this function will be negative, so we're going to get divergence to $-\infty$ as $x \to x_1^+$ and as $x \to x_2^-$. In between, we're going to have to get some kind of local maximum. This can be computed by finding the minimum of the denominator (which in our case, invariably occurs at $x = -1$).
When $x$ is not between $x_1$ and $x_2$, the function is positive, and will range from $0$ (the horizontal asymptote) to $\infty$ (the vertical asymptote) and take every value in between. So, our range will take the form $$(-\infty, m] \cup (0, \infty)$$ where $m$ is the local maximum (the function value at $1$).
Using the discriminant of $x^2 + 2x + 4a - 2$ we get distinct roots when $$4 - 4(4a - 2) > 0 \iff a < \frac{3}{4},$$ and we have $m = \frac{1}{4a-3}$, thus our range is $$\left(-\infty, \frac{1}{4a-3}\right] \cup (0, \infty).$$ When we have a repeated root, i.e. when $a = \frac{3}{4}$, then our function is $$\frac{1}{(x + 1)^2}.$$ The negative bit of the function is now gone, and on both sides of the one asymptote, the function tends to $\infty$. Our range becomes $$(0, \infty).$$ Otherwise, if $a > \frac{3}{4}$, we still get $0$ convergence as $x \to \pm \infty$, but now we are positive and continuous everywhere. Therefore, we must achieve a maximum somewhere, particularly where the parabola $x^2 + 2x + 4a - 2$ achieves its minimum. This occurs at $x = -1$ again, so our range is $$\left(0, \frac{1}{4a - 3}\right].$$
As you can see, this is a fairly ad-hoc process. Unfortunately, there isn't really a way around this. All the tools that people use for graphing functions also come in handy in figuring out their ranges, such as identifying asymptotes, local maxima/minima, etc. All I can say is, good luck.