# How to find inverse of a composite function?

I am stuck with this question,

Let $A=B=C=\mathbb{R}$ and consider the functions $f\colon A\to B$ and $g\colon B\to C$ defined by $f(a)=2a+1$, $g(b)=b/3$. Verify Theorem 3(b): $(g\circ f)^{-1}=f^{-1}\circ g^{-1}.$

I have calculated $f^{-1}$, $g^{-1}$, and their composition, but how do I find the inverse of $(g\circ f)$?

Here is how I have done so far,

\begin{align*} \text{Let}\qquad\qquad b &= f(a)\\ a&= f^{-1}(b)\\ &{ }\\ b&=f(a)\\ b&=2a+1\\ \frac{b-1}{2} &= a\\ a &= \frac{b-1}{2} \end{align*} But $a=f^{-1}(b)$, $$f^{-1}(b) = \frac{b-1}{2}.$$

${}$

\begin{align*} \text{Let}\qquad\qquad a&=g(b)\\ b&= g^{-1}(a)\\ a&= g(b)\\ a &= b/3\\ b &= 3a\\ g^{-1}(a) &= 3a\qquad(\text{because }b=g^{-1}(a) \end{align*}

${}$

\begin{align*} f^{-1}\circ g^{-1} &= ?\\ &= f^{-1}\Bigl( g^{-1}(a)\Bigr)\\ &= f^{-1}(3a)\\ f^{-1}\circ g^{-1} &= \frac{3a-1}{2} \end{align*}

\begin{align*} g\circ f&= g\bigl(f(a)\bigr)\\ &= g(2a+1)\\ g\circ f &= \frac{2a+1}{3}\\ (g\circ f)^{-1} &= ?\\ \text{Let}\qquad\qquad &b=g\circ f \end{align*}

EDIT:

Thanks for the answers, I followed the suggestions and came up with the answer, Now I have two questions,

1. The answers do match but the arguments are different. Is that ok?
2. Is $(g\circ f)$ same as $(g\circ f(a))$?
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$(g\circ f)$ is a function. $(g\circ f(a))$ is the value of the function $g\circ f$ at $a$. They are not the same thing (one is a function, the other is an number that you've written in parentheses). The name of the variable doesn't matter. The function $g(x)=x^2$ is the same as the function $g(z)=z^2$. – Arturo Magidin Aug 26 '11 at 16:46

$f,g$ are the functions defined in the question.

We have

$$\begin{eqnarray*} b &=&f(a)=2a+1, \end{eqnarray*}$$

or equivalently, by definition of the inverse function $f^{-1}$

$$\begin{eqnarray*} &a=\frac{b-1}{2}=f^{-1}(b).\tag{A} \end{eqnarray*}$$

Since

$$\begin{eqnarray*} c &=&g(b)=\frac{b}{3}, \end{eqnarray*}$$

or equivalently, by definition of the inverse function $g^{-1}$

$$\begin{eqnarray*} b=3c=g^{-1}(c),\tag{B} \end{eqnarray*}$$

after combining $(A)$ and $(B)$, we get

$$a=\frac{3c-1}{2}=(f^{-1}\circ g^{-1})(c).\tag{1}$$

On the other hand

$$c=(g\circ f)(a)=g(f(a))=g(2a+1)=\frac{2a+1}{3}.\tag{2}$$

Hence, by definition, the value at $c$ of the inverse function $(g\circ f)^{-1}$, is

$$a=\frac{3c-1}{2}=(g\circ f)^{-1}(c).\tag{3}$$

From $(1)$ and $(3)$ we conclude that for these functions $f,g$ and their inverses $f^{-1},g^{-1}$ the following identity holds:

$$(f^{-1}\circ g^{-1})(c)=(g\circ f)^{-1}(c).\tag{4}$$

Notation's note: $(f^{-1}\circ g^{-1})(c)=f^{-1}(g^{-1}(c))$.

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Thanks a lot for such an effort. – Fahad Uddin Aug 26 '11 at 19:25
@fahad: You are welcome. – Américo Tavares Aug 26 '11 at 19:28
@AméricoTavares I usually draw the function in this way eevry time it helps me with composition of functions and when I "play" with algebraic structures but I was always scared to use them in order to explain my concepts in my questions to other people because I believed I was taking too much freedom with a notation that I only saw in cathegory theory, where your sets A,B and C are usually set with sturctures. So is correct to use these "diagrams" even outside the cathegory theory context? – MphLee Apr 20 '13 at 16:25
@MphLee I am not a mathematician and know nothing about cathegory theory, but these diagrams appeared in some books of Calculus/Real Analysis for Engineers back in 1060-1970's. – Américo Tavares Apr 20 '13 at 17:03
@AméricoTavares and I'm even less a mathematican I'll try to search more about this and I'll continue use these diagrams in my amatorial practice, thanks anyways – MphLee Apr 20 '13 at 17:44

You find the inverse of $g\circ f$ by using the fact that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$.

In other words, what gets done last gets undone first.

$f$ multiplies by 2 and then adds 1.

$g$ divides by 3.

Dividing by 3 is done last, so it's undone first.

The inverse first multiplies by 3, then undoes $f$.

Later note: Per the comment, to verify that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$:

Instead of confusingly writing $a = g(b)$, write $c=g(b)$. Then $c=b/3$, so $b=3c$, so $$g^{-1}(c) = 3c.$$ And $$f^{-1}(b) = \frac{b-1}{2}.$$

So $$b = 3c\qquad\text{and}\qquad a = \frac{b-1}{2}.$$ Put $3c$ where $b$ is and get $$a=\frac{3c-1}{2}.$$ You want to show that that's the same as what you'd get by finding $g(f(a))$ directly and then inverting.

So $c = g(f(a)) = \dfrac{f(a)}{3} = \dfrac{2a+1}{3}$.

So take $c = \dfrac{2a+1}{3}$ and solve it for $a$:

\begin{align} 3c & = 2a+1 \\ 3c - 1 & = 2a \\ \\ \frac{3c-1}{2} & = a. \end{align}

FINALLY, observe that you got the same thing both ways.

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But the problem asks the student to verify the formula; that is, find the inverse of $g\circ f$ "directly", and then compare it to the function you get by computing $f^{-1}\circ g^{-1}$. Surely using the formula to verify that the formula works is a tad... unsatisfying. – Arturo Magidin Aug 26 '11 at 15:58
I am stuck with how do I find (gof)^-1 – Fahad Uddin Aug 26 '11 at 16:14
OK, I've added a later note. – Michael Hardy Aug 26 '11 at 16:18
@MichaelHardy thank you for this answer, it came in handy for a question I recently asked – seeker Dec 15 '13 at 13:01

What you've done so far is to compute $f^{-1}$ and $g^{-1}$, and $f^{-1}\circ g^{-1}$. Now you want to try to find $(g\circ f)^{-1}$ directly, and compare that to what you've computed (in order to verify the formula).

So, you've figured out that $(g\circ f)(a) = \frac{2a+1}{3}$. How do we figure out $(g\circ f)^{-1}$?

Exactly the same way we figure out the inverse of any function. If someone stopped you on the street, pointed a gun at you and said

"Here, I have this function: $$h(a) = \frac{2a+1}{3},$$ I need the formula for $h^{-1}$. Give it to me or I'll shoot you!"

then you don't need to know where that function came from, all you need to do is figure out the inverse: \begin{align*} b &= \frac{2a+1}{3}\\ 3b &= 2a+1\\ 3b-1 &= 2a\\ &\vdots \end{align*} etc. When you are done and have a formula for $h^{-1}(a) = (g\circ f)^{-1}(a)$, you can compare it to the formula you found for $f^{-1}\circ g^{-1}$ and verify that you got the same function.

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Thanks alot. I did it as you suggested. A small problem that I am having is that is in terms of b instead of a. Checkout the edit. – Fahad Uddin Aug 26 '11 at 16:30
@fahad: Don't get hung up on the letters. The name of the variable is immaterial. The function $h(x) = x^2$ is exactly the same as the function $h(y)=y^2$, which is exactly the same as the function $h(z)=z^2$, which is exactly the same as the function $h(a)=a^2$. Just switch all the $b$s into $a$s and be done. – Arturo Magidin Aug 26 '11 at 16:36
Thanks alot for the answer :) – Fahad Uddin Aug 26 '11 at 19:15
I can't resist adding that the Russian mathematical physicist Igor Tamm was once told to do a mathematics calculation or be shot. :) – Mike Spivey Aug 26 '11 at 22:21