Inverse function question. Problem with point of intersection. (1 Viewer)

DuhBoyChong · Apr 29, 2015

Hello, im having trouble with sketching the inverse function of f(x)= tan(x/2), -π≤x≤π .

I have the domain, range, period, and phase. I have drawn the mirror y=x, but i can't find the point of intersection. in which i equate y=x, and y=tan(x/2) to find where the mirror line and the forward function intersects. How do you simultaneously solve y=x, and y=tan(x/2),

It it x=tan(x/2)? Dunno what to do.

integral95 · Apr 29, 2015

Are you just required to sketch the inverse function, in that case you just need the equation of the inverse equation and not worry about the point of intersection with y = x

$$The inverse of$ f(x) = tan(\frac{x}{2}) \ \ \ \is \\ \\ \\ f^{-1}(x) = 2tan^{-1}(x) , $since the domain of the function is$ \\ \\ -\pi \leq x \leq \pi \\ \\ \\ $ The range of the inverse function is$ \ \ \ -\pi \leq y \leq \pi$

DuhBoyChong · Apr 29, 2015

integral95 said:
Are you just required to sketch the inverse function, in that case you just need the equation of the inverse equation and not worry about the point of intersection with y = x

$$The inverse of$ f(x) = tan(\frac{x}{2}) \ \ \ \is \\ \\ \\ f^{-1}(x) = 2tan^{-1}(x) , $since the domain of the function is$ \\ \\ -\pi \leq x \leq \pi \\ \\ \\ $ The range of the inverse function is$ \ \ \ -\pi \leq y \leq \pi$

Yea, sorry, the domain was -π≤x≤π and thanks.

InteGrand · Apr 29, 2015

By the way, those should be strict inequalities for the (natural) domain ( $-\pi < x < \pi$ ), since the tan function would be undefined at the boundaries.

Inverse function question. Problem with point of intersection. (1 Viewer)

DuhBoyChong

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integral95

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DuhBoyChong

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InteGrand

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