If $f(x)=g(h(x))$, why is $f^{-1}(x)=h^{-1}(g^{-1}(x))$ ?
A question about composition of functions
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2I am trying to show my calculus class how to use stackexchange. – 2012-09-02
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0Do you mean that you intend to answer the question yourself? – 2012-09-02
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0Only if absolutely necessary :) – 2012-09-02
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1I suppose that you are talking about pre image (inverse image), not about inverse function! – 2012-09-02
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3If getting dressed means first to put on briefs and then trousers, then getting undressed means to first take off the trousers and then the briefs. – 2012-09-02
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0The question as posed is exactly as it was asked to me. f^{-1}(x) is the inverse function to f(x). – 2012-09-02
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0What gets done last gets undone first. See my answer below. (And tell them all to vote for it :-) ). – 2012-09-02
4 Answers
Think about dressing your feet. Here are the instructions
- Put on socks
- Put on shoes
What is the reverse of this operation?
- Remove shoes
- Remove socks.
You must undo the operations in the reverse order in which you did them.
Now think about f(g(x)): first apply g to x then f.....
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0Simple! Simple! Simple!. All can undrestand what to do here. – 2012-09-02
What gets done last gets undone first. Thus: $$ \begin{array}{c} \text{input} & \mapsto & \text{multiply by }5 & \mapsto& \text{add }2 \\[10pt] x & \mapsto & 5x & \mapsto & 5x+2 = y \end{array} $$ The inverse is: $$ \begin{array}{c} \text{input} & \mapsto & \text{subtract }2 & \mapsto & \text{divide by }5 \\[10pt] y & \mapsto & y-2 & \mapsto & \frac{y-2}{5} \end{array} $$
By definition, $f^{-1}$ is the function with the following properties: for each $x$ in the domain of $f$, $f^{-1}(f(x))=x$, and for each $x$ in the range of $f$, $f(f^{-1}(x))=x$. In other words, $f^{-1}$ undoes the effects of $f$, and $f$ undoes the effects of $f^{-1}$.
If $f(x)=g(h(x))$, then in order to undo the effects of $f$ you have to undo those of $g$ to get at $h(x)$, and then you have to undo those of $h$ to get at $x$. In other words, you must first apply $g^{-1}$ to $f(x)$, and then you must apply $g^{-1}$ to the result. This is actually easier to follow in symbols than in words:
$$\begin{align*} h^{-1}\Big(g^{-1}\big(f(x)\big)\Big)&=h^{-1}\left(g^{-1}\Big(g\big(h(x)\big)\Big)\right)\\ &\overset{(*)}=h^{-1}\big(h(x)\big)\\ &=x\;. \end{align*}$$
The starred step makes use of the fact that $g^{-1}\big(g(u)\big)=u$ no matter what $u$ is, provided that it’s in the domain of $g$.
The calculation showing that $g\left(h\Big(h^{-1}\big(g^{-1}(x)\big)\Big)\right)=x$ is entirely similar.
If you take each function as a transformation, $g(h(x))$ means first apply the transformation described by $h(x)$ to $x$, then apply the transformation described by $g(x)$ to the result of that.
Therefore, in order to get the original image from the result of these transformations, we first undo the latest transformation (in this case $g(x)$), and then undo the transformation described by $h(x)$.
Therefore, we have $f^{-1}(x)=h^{-1}(g^{-1}(x))$.