# Proof for invertible function

I'm trying to prove that my function is invertible. The function is:

  f(x) =     4x-9       when x > 3


I have drawn the graph and know that its "flipped" version is its invertible function.

The inverse function of:

f(x) = 4x-9        when x > 3


I would assume is:

f^−1(x) = x/4 + 3   when x > 3


The graph makes clear that an inverse exists and for $$y$$ you must discern the cases $$y<4$$, $$4\leq y\leq 6$$ and $$y>6$$.

• $$y<4$$ Then equation $$y=f(x)$$ can be rewritten as $$y=4x-8$$ and results in $$x=\frac14y+2$$.
• $$4\leq y\leq6$$ Then equation $$y=f(x)$$ can be rewritten as $$y=-\frac23x+8$$ and results in $$x=12-\frac32y$$.
• $$y\geq6$$ Then equation $$y=f(x)$$ can be rewritten as $$y=4x-18$$ and results in $$x=\frac14y+\frac92$$.

Switching the roles of $$x$$ and $$y$$ we arrive at inversion:

$$g\left(x\right)=\begin{cases} \frac{1}{4}x+2 & \text{if }x<4\\ 12-\frac{3}{2}x & \text{if }4\leq x\leq6\\ \frac{1}{4}x+\frac{9}{2} & \text{if }x>6 \end{cases}$$

This was done on base of the graph, but a formal proof that $$f$$ and $$g$$ are indeed inverses of each other can now be given by showing that the compositions $$f\circ g$$ and $$g\circ f$$ both coincide with the identity function on $$\mathbb R$$.

• Okay seems reasonable. But why is it x < 4 instead of x < 3 like it was in the beginning? – Blue shirt Nov 21 '19 at 10:37
• Because the role of $x$ is now taken over by $y$ you could say. Look at the graph and see that the important points are $(3,4)$,$(3,6)$, $(6,4)$ and $(6,6)$. For the inverse these points turn into $(4,3)$,$(6,3)$,$(4,6)$ and $(6,6)$ by switching the coördinates. – drhab Nov 21 '19 at 10:42