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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

user639649
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  • Actually $f(x)$ is only positive if $x$ is positive. Have you considered what happens if $x < 0$? – Minus One-Twelfth Apr 18 '19 at 21:37
  • @MinusOne-Twelfth if x < 0 then doesn't one part of the curve continue to tend towards 0? – user639649 Apr 18 '19 at 21:41
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    The range won't be $f(x)\le -1/4$, because of course $f(x)$ isn't always less than or equal to $-1/4$ (e.g. $f(x)$ can be positive too). Did you mean $\color{red}{\ge}$? Anyway, here is the graph of the function. You can see from there that the range is $f(x)\ge -1/4$. – Minus One-Twelfth Apr 18 '19 at 21:42
  • The textbook says ≤ , which is why I'm confused @MinusOne-Twelfth – user639649 Apr 18 '19 at 21:44

3 Answers3

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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!

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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 $A = \{ y: y \in \mathbb{R}, y \ge -\frac{1}{4}\}$.

DeepSea
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$(1+x)^{2} \geq 0$ 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 $[-\frac 1 4, \infty)$.

Kavi Rama Murthy
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