Range of $\frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$? I have the following problem:
Let $f(x) = \frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$ with $c>0$ and $x \in \mathbb{R}$. How to prove that $0 \leq f(x) \leq 1$?
I'm not sure how to check the range, I tried plotting the function but that's not as formal as the problem asks. Though, at least I'm sure that the range is correct. Thanks for the help.
 A: First to check if $0 \leq f(x)$ everywhere:  $c>0,$ anything squared is $\geq 0$, so yes.  The minimum is reached at $x=0$ only, because the numerator of a fraction must be $0$ for the fraction to be $0$.
If $c=1$ then the function is $0$ everywhere.  If $c$ is large we have roughly $c^3/c^4 = 1/c$ so that will be $\leq 1$.  We could try small values of $c$, but I don't see any values of $x$ that will bring the function over $1$.
The calculus to check for extrema expanded out is straightforward but messy: $21$ terms.  So we simplify by pulling out constants and expand the denominator to get $$\frac{x^2}{x^4 + x^2(c^3+c) + c^4}$$ and then we set $y=x^2$ and relabel to get
$$\frac{y}{y^2+ay+b}$$ which is far more tractable, and has extrema at $y^2 = b$.  Substituting $x^2 = y$ and $c^4 = b$, we get $x = \pm c$ as locations of the extrema.
For the values there, we get $$\frac{(1-c)^2}{(1+c)^2} = \frac{(c-1)^2}{(1+c)^2}$$ which is less than $1$ for all $c>0$.
A: You don't even need calculus:
$$f(x) = \dfrac{x^2c(1-c)^2}{x^4 + x^2c(c^2+1)+c^4}\leq\dfrac{x^2c(1-c)^2}{x^2c(c^2+1)} = \dfrac{c^2-2c+1}{c^2+1}\leq 1$$
assuming non-negative $c.$
