Find the range of $\frac{x^2}{2(1-\cos x)}$ for $0

Question

Find the range of $$f(x) =\frac{x^2}{2(1-\cos x)}$$ for $$0<x<1$$

I tried to solve by writing $x$ in terms of $y,$ but no result.

Quick beginner guide for asking a well-received question + please avoid "no clue" questions. — Anne Bauval, Jul 16 '23 at 18:06
You can see the function monotonically increases if you check its derivative — Kay K., Jul 16 '23 at 18:11
Someone has noted that you can see it is monotonically increasing if you take a derivative, if you are not familiar with derivatives you could also note that it can be expressed as the product of two monotonically increasing functions, $$ x^2 $$ and $$ \frac{1}{1-cos(x)} $$ so it is itself a monotonically increasing function — Carlyle, Jul 16 '23 at 18:12
@AnneBauval you are right. For x =$2\pi{n}$ where $n\in{Z}$. Function has undefined values. — , Jul 16 '23 at 18:21
@MaximusFastidiousIrreverence L'Hopital is but one of the many ways to prove this limit. — Anne Bauval, Jul 16 '23 at 21:22
$f'(x)=\frac{xg(x)}{2\left(1-\cos x\right)^2}$ where $g(x)=2\left(1-\cos x\right)-x\sin x.$
From $g(0)=0$ and $\forall x\in(0,\pi/2)\quad g'(x)=\cos x\left(\tan x-x\right)>0,$ we derive that $f$ is increasing on its domain $(0,1).$ By continuity, its range is therefore $$(\lim_0f,\lim_1f)=\left(1,\frac1{2\left(1-\cos 1\right)}\right).$$ — Anne Bauval, Jul 16 '23 at 21:47

score -1 · Answer 1 · answered Jul 16 '23 at 22:48

-1

You can prove that this function is monotonic at this range, and then calculate:
$f(1)=\frac{1}{2(1-cos(1))}, \lim_{x\to 0^+}f(x)=1$, and so this is the range.

answered Jul 16 '23 at 22:48

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Find the range of $\frac{x^2}{2(1-\cos x)}$ for $0

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