# Maximal value of domain for a function by looking at inverse function.

The function g:[–a,a]→ R, g(x)=sin(2(x-π/6))has an inverse function.The maximum possible value of a is:

From what I understand the domain of g(x) is the range of g'(x). So I would try to find the inverse equation, work out the range, and then use this information to find the value of a, in the domain.

but i'm stuck at finding the range.

note: if i enter π/2 the calc says undefined so i'm assuming the value of a is π/2.

the answer is out of: π/12, 1, π/6, π/4, π/2

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– user61527
Commented Jan 2, 2014 at 4:46

$$\sin(2(x-\pi/6))=\sin(2x - \pi/3)$$

$\sin$ has an inverse in the interval $-\pi/2$ to $\pi/2$. So $$-\pi/2 \le 2x-\pi/3 \le \pi/2 \\ -\pi/2+\pi/3\le 2x \le \pi/2 + \pi/3 \\ -\pi/12 \le x \le 5\pi/12$$ So $[-a, a]$ should fit into $[-\pi/12, 5\pi/12]$ So $$a = \pi/12$$

The following was my original answer and is wrong! Please disregard. I was not thinking right!

Note that $\sin x$ has an inverse in an interval of length $\pi$. So $\sin 2x$ must have an inverse over an interval of length $\pi/2$. So $a=\pi/4$.

• I'm not really understanding what an interval length is. Commented Jan 2, 2014 at 5:25
• See my edited answer Commented Jan 2, 2014 at 5:40
• Uh... where? is it? Commented Jan 2, 2014 at 5:42
• sorry, I am a slow typist :) Commented Jan 2, 2014 at 5:47
• i dont understand how you get pi/6 on both sides? Can you explain please? haha, it's okay, I'm slower :( Commented Jan 2, 2014 at 5:53