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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.

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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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  • $\begingroup$ You may find this guide to writing math useful in formatting your question. $\endgroup$
    – user61527
    Commented Jan 2, 2014 at 4:46

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

$\sin$ 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 $a=\pi/4$.

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  • $\begingroup$ I'm not really understanding what an interval length is. $\endgroup$
    – confused
    Commented Jan 2, 2014 at 5:25
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    $\begingroup$ See my edited answer $\endgroup$
    – user44197
    Commented Jan 2, 2014 at 5:40
  • $\begingroup$ Uh... where? is it? $\endgroup$
    – confused
    Commented Jan 2, 2014 at 5:42
  • $\begingroup$ sorry, I am a slow typist :) $\endgroup$
    – user44197
    Commented Jan 2, 2014 at 5:47
  • $\begingroup$ i dont understand how you get pi/6 on both sides? Can you explain please? haha, it's okay, I'm slower :( $\endgroup$
    – confused
    Commented Jan 2, 2014 at 5:53

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