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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
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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 $\mathbb R$.

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  • $\begingroup$ Okay seems reasonable. But why is it x < 4 instead of x < 3 like it was in the beginning? $\endgroup$ – Blue shirt Nov 21 '19 at 10:37
  • $\begingroup$ 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. $\endgroup$ – drhab Nov 21 '19 at 10:42

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