# How do you find f^-1x and how must x be restricted in f^-1x given f(x)=4+2cos(x-3) and 3<=x<=3+pi?

Jan 31, 2018

As with all inverse functions, start by taking equation $y = f \left(x\right)$ and substituting $x = f \left(y\right)$

We have $y = f \left(x\right) = 4 + 2 \cos \left(x - 3\right)$, hence we let: $x = 4 + 2 \cos \left(y - 3\right)$

Doing some algebraic manipulation we get:

$\frac{1}{2} \left(x - 4\right) = \cos \left(y - 3\right)$
$\therefore y - 3 = {\cos}^{-} 1 \left(\frac{1}{2} \left(x - 4\right)\right)$
$\therefore y = {\cos}^{-} 1 \left(\frac{1}{2} \left(x - 4\right)\right) + 3$

This is our ${f}^{-} 1$, but first we must find the new domain.

Since $\mathrm{do} m f = \left[3 , 3 + \pi\right]$ and the cosine function is one-to-one within any interval of $\pi$, there is no need to restrict further. Now $\mathrm{do} m f$ becomes $r a n {f}^{-} 1$.

The implied domain of ${f}^{-} 1$ gives $\frac{1}{2} \left(x - 4\right) \in \left[0 , \pi\right]$, which implies that $x \in \left[4 , 2 \pi + 4\right]$

Hence:

${f}^{-} 1 : \left[4 , 2 \pi + 4\right] \rightarrow \mathbb{R} , {f}^{-} 1 \left(x\right) = {\cos}^{-} 1 \left(\frac{1}{2} \left(x - 4\right)\right) + 3$