Find $f^{-1}(x)$ when $\\simplify{f(x)={a[0]}x+{a[1]} }$.
\n$\\displaystyle f^{-1}(x)=$ [[0]]
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\n$\\displaystyle g^{-1}(x)=$ [[0]]
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$\\displaystyle (g^{-1}\\circ f^{-1}) (x)$ =[[0]]
Using:
\\[
\\begin{align}
f(x)&=\\simplify{{a[0]}x+{a[1]} }\\\\
&\\text{ and } \\\\
g(x)&=\\simplify{{a[3]}x-{a[2]} }\\text{,}
\\end{align}
\\]
find $(f\\circ g)(x)$, the composition of $f(x)$ with $g(x)$.
$\\displaystyle (f\\circ g)(x)=$ [[0]]
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\n$\\displaystyle (f\\circ g)^{-1}(x)=$ [[0]]
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\nTo find this, we first set $x=f(y)$ and rearrange to find $y$ in terms of $x$, i.e. $y = f^{-1}(x)$.
\n\\begin{align}
f(y)=\\simplify{{a[0]}y+{a[1]}}&=x\\\\
\\simplify{{a[0]}y}&=x-\\var{a[1]}\\\\[1em]
y&=\\simplify[]{(x-{a[1]})/{a[0]}}\\\\[1em]
f^{-1}(x)&=\\simplify{(x-{a[1]})/{a[0]}}\\text{.}\\\\
\\end{align}
We use the same method as part a) to find $g^{-1}(x)$:
\n\\begin{align}
g(y)=\\simplify{{a[3]}y-{a[2]}}&=x\\\\
\\simplify{{a[3]}y}&=x+\\var{a[2]}\\\\[1em]
y&=\\simplify[]{(x+{a[2]})/{a[3]}}\\\\[1em]
g^{-1}(x)&=\\simplify{(x+{a[2]})/{a[3]}}\\text{.}\\\\
\\end{align}
$(g^{-1} \\circ f^{-1})(x)$ is the function which first applies $f^{-1}(x)$ and then applies $g^{-1}$ to the result of that.
\nWe use the previous answers: $f^{-1}(x)=\\simplify{(x-{a[1]})/{a[0]}}$ and $g^{-1}(x)=\\simplify{(x+{a[2]})/{a[3]}}$ to find the definition of $(g^{-1} \\circ f^{-1})(x)$ by substituting $f^{-1}(x)$ everywhere $x$ occurs in the definition of $g^{-1}(x)$.
\n\\begin{align}
(g^{-1}\\circ f^{-1}) (x)&=g^{-1}(f^{-1}(x))\\\\[1em]
&=g^{-1} \\left( \\simplify[]{(x-{a[1]})/{a[0]}} \\right) \\\\[1em]
&=\\frac{\\left(\\simplify[]{(x-{a[1]})/{a[0]}}\\right)+\\simplify[]{{a[2]}}}{\\var{a[3]}\\text{.}}
\\end{align}
$(f \\circ g)(x)$ is the function which first applies $g(x)$ and then applies $f$ to the result of that.
\nWe find the definition of $(f \\circ g)(x)$ by substituting $g(x)$ everywhere that $x$ occurs in the definition of $f(x)$.
\n\\begin{align}
(f\\circ g)(x)&=f(g(x))\\\\
&=f(\\simplify{{a[3]}x-{a[2]}})\\\\
&=\\simplify{{a[0]}({a[3]}x-{a[2]})+{a[1]}}
\\end{align}
Now that we have the definition of $(f \\circ g)(x)$, we can find its inverse by using the same method as in parts a) and b).
\n\\begin{align}
(f \\circ g)(y) &= x \\\\
\\simplify{{a[0]}({a[3]}y-{a[2]})+{a[1]}}&=x\\\\
\\simplify{{a[0]}({a[3]}y-{a[2]})}&=x-\\var{a[1]}\\\\[1em]
\\simplify{{a[3]}y-{a[2]}}&=\\frac{(x-\\var{a[1]})}{\\var{a[0]}}\\\\[1em]
\\simplify{{a[3]}y}&=\\left( \\frac{(x-\\var{a[1]})}{\\var{a[0]}} \\right) +\\var{a[2]}\\\\[1em]
y&=\\frac{\\left(\\frac{(x-\\var{a[1]})}{\\var{a[0]}}\\right)+\\var{a[2]})}{\\var{a[3]}}\\\\[1em]
(f\\circ g)^{-1}(x)&=\\frac{\\left(\\frac{(x-\\var{a[1]})}{\\var{a[0]}}\\right)+\\var{a[2]}}{\\var{a[3]}}\\\\[1em]
\\end{align}
We can see that in this case $(f\\circ g)^{-1}(x) = (g^{-1}\\circ f^{-1}) (x)$.
\nSo long as the inverses of $f$ and $g$ exist and they can be composed, it is always true that \\[(f \\circ g)^{-1}(x) \\equiv (g^{-1} \\circ f^{-1}) (x)\\text{.}\\]
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\nFunction composition is applying one function to the results of another.
\nThe following questions will ask you to find the inverses and compositions of some functions.
\nGive all of your answers in terms of $x$.
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