# Inverse Functions: Introduction

Inverse functions are a pair of function that perform the opposite operations. The inverse function of f(x) is denoted by f-1(x), read "f-inverse". For example f(x) = x - 2 has an inverse function f-1(x) = x + 2 because for any value of x the value for f(x) when substituted into f-1(x) equals x.

$f\left[{f}^{-1}\left(x\right)\right]=f\left(x+2\right)=\left(x+2\right)-2=x$

Another way to describe inverse functions is to say that the set of ordered pairs for f(x) is the opposite of the set of ordered pairs for f-1(x).

f(x) = x - 2:      (1, -1), (2, 0), (3, 1), (4, 2)

f-1(x) = x + 2:      (-1, 1), (0, 2), (1, 3), (2, 4)

DEFINITION OF AN INVERSE FUNCTION:

An inverse function is a function such that:

$f\left[{f}^{-1}\left(x\right)\right]=x$        and       ${f}^{-1}\left[f\left(x\right)\right]=x$

Where the domain of f(x) equals the range of f -1 (x) and the range of f(x) equals the domain of f -1 (x)

Let's use this definition to verify inverse functions.

Example 1: Show that f(x) = 2x + 1 and $g\left(x\right)=\frac{x-1}{2}$ are inverse functions.
 Step 1: Show that $f\left[g\left(x\right)\right]=x$. Substitute for g(x) $f\left[g\left(x\right)\right]=f\left(\frac{x-1}{2}\right)$ Substitute for the expression in f(x) $f\left(\frac{x-1}{2}\right)=2\left(\frac{x-1}{2}\right)+1$ Solve $2\left(\frac{x-1}{2}\right)+1=\left(x-1\right)+1=x$ Step 2: Show that $g\left[f\left(x\right)\right]=x$. Substitute for f(x) $g\left[f\left(x\right)\right]=g\left(2x+1\right)$ Substitute the expression in g(x) $g\left(2x+1\right)=\frac{\left(2x+1\right)-1}{2}$ Solve $\frac{\left(2x+1\right)-1}{2}=\frac{2x}{2}=x$ Step 3: Show that the domain/range of f(x) is equal to the range/domain of g(x). Substitute values for x into f(x) then substitute the values obtained for f(x) into g(x) and compare. Because the domain/range of f(x) is equal to the range/domain of g(x), f(x) and g(x) are inverse functions.
Example 2: Show that $f\left(x\right)=\frac{{x}^{3}}{27}$ and $g\left(x\right)=\sqrt{27x}$ are inverse functions.
 Step 1: Show that $f\left[g\left(x\right)\right]=x$. Substitute for g(x) $f\left[g\left(x\right)\right]=f\left(\sqrt{27x}\right)$ Substitute the expression in f(x) $f\left(\sqrt{27x}\right)=\frac{{\left(\sqrt{27x}\right)}^{3}}{27}$ Solve $\frac{{\left(\sqrt{27x}\right)}^{3}}{27}=\frac{27x}{27}=x$ Step 2: Show that $g\left[f\left(x\right)\right]=x$. Substitute for f(x) $g\left[f\left(x\right)\right]=g\left(\frac{{x}^{3}}{27}\right)$ Substitute the expression in g(x) $g\left(\frac{{x}^{3}}{27}\right)=\sqrt{27\left(\frac{{x}^{3}}{27}\right)}$ Solve $\sqrt{27\left(\frac{{x}^{3}}{27}\right)}=\sqrt{{x}^{3}}=x$ Step 3: Show that the domain/range of f(x) is equal to the range/domain of g(x). Substitute values for x into f(x) then substitute the values obtained for f(x) into g(x) and compare. Because the domain/range of f(x) is equal to the range/domain of g(x), f(x) and g(x) are inverse functions.

 Related Links: Math algebra Inverse Functions: Graphs Inverse Functions: One to One Algebra Topics

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