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If $f(x) = \\dfrac{\\simplify{{m}x + {c}}}{\\simplify{x + {a}}}$, $x \\neq \\simplify{{-a}}$, find the inverse function $f^{-1}(x)$ v\u1edbi $c = \\var{c}, a = \\var{a}, m = \\var{m}$

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To find $f^{-1}x$, it can help to first set $f(x)$ to a different variable, which we will call $y$

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$y = f(x) = \\dfrac{\\simplify{{m}x + {c}}}{\\simplify{x + {a}}}$

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Since the function $f(x)$ take us from $x$ to $y$, the inverse function $f^{-1}$ will take us from $y$ to $x$. So to obtain $f^{-1}$, we want to rearrange $y = f(x) = \\dfrac{\\simplify{{m}x + {c}}}{\\simplify{x + {a}}}$ so that it's $x$ as a function of $y$.

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$y = \\dfrac{\\simplify{{m}x + {c}}}{\\simplify{x + {a}}}$

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$\\simplify{(x + {a})y} = \\simplify{{m}x + {c}}$

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$\\simplify{xy + {a}y} = \\simplify{{m}x + {c}}$

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$\\simplify{xy - {m}x} = \\simplify{{c} - {a}y}$

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$\\simplify{x(y - {m})} = \\simplify{{c} - {a}y}$

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$x = \\dfrac{\\simplify{{c} - {a}y}}{\\simplify{y - {m}}}$

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Hence, $f^{-1}(y) = \\dfrac{\\var{c} - \\var{a}y}{y - (\\var{m})}$, and therefore

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$f^{-1}(x) = \\dfrac{\\simplify{{c} - {a}y}}{\\simplify{y - {m}}}$

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(Note: The inverse is valid for all values of $x$ except $x = \\var{m}$, since this would make the denominator equal to 0.)

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$f^{-1}(x) = $ [[0]]  

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