Finding the range of a reciprocal quadratic graph

Question

How would you find the range of this graph?

$$f(x) = \frac{x}{(1-x)^2}$$

I understand that $f(x)>0$ but the range is actually $f(x)\le -\frac{1}{4}$ but I don't understand why? I've attempted to draw the graph and still don't understand.

Thank you

Answer 1

I think the correct sign for the range should be $f(x)\ge-\frac{1}{4}$

First you can plot the graph. Desmos is great!

But we can find this answer without it by finding the absolute minimum.

If we differentiate $f(x)$ we get $f'(x)=\frac{1+x}{(1-x)^3}$, then set this equal to zero and solve to get $x=-1$ so plug this back into the equation and we get $f(x)=-0.25$

We can then see if it is a minimum point by looking the the behaviour of $x$ near $x=-1$! Hope this helps!

Answer 2

Put $y = \dfrac{x}{(1-x)^2}\implies y(x^2-2x+1) = x \implies yx^2 -(2y+1)x + y = 0$. This equation has solution in $x$ meaning that $\triangle \ge 0\implies (2y+1)^2 - 4y^2\ge 0\implies y \ge -\dfrac{1}{4}$, and this means the range is the set $A = \{ y: y \in \mathbb{R}, y \ge -\frac{1}{4}\}$.

Answer 3

$(1+x)^{2} \geq 0$ so $(1-x)^{2} \geq -4x$. This gives $f(x) \geq -\frac 1 4$. If $y \geq -\frac 1 4$ you can solve a quadratic to find $x$ such that $f(x)=y$. I leave this calculation to you. So the range is $[-\frac 1 4, \infty)$.