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.

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

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:

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

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 $g (x) = \frac{x - 1}{2}$ are inverse functions.

Step 1: Show that $f [g (x)] = x$ .

Substitute for g(x)

$f [g (x)] = f (\frac{x - 1}{2})$

Substitute for the expression in f(x)

$f (\frac{x - 1}{2}) = 2 (\frac{x - 1}{2}) + 1$

Solve

$2 (\frac{x - 1}{2}) + 1 = (x - 1) + 1 = x$

Step 2: Show that $g [f (x)] = x$ .

Substitute for f(x)

$g [f (x)] = g (2 x + 1)$

Substitute the expression in g(x)

$g (2 x + 1) = \frac{(2 x + 1) - 1}{2}$

Solve

$\frac{(2 x + 1) - 1}{2} = \frac{2 x}{2} = x$

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.

$f (x) = 2 x + 1 : (0, 1), (1, 3), (2, 5), (3, 7)$

$g (x) = 2 x + 1 : (1, 0), (3, 1), (5, 2), (7, 3)$

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 $f (x) = \frac{x^{3}}{27}$ and $g (x) = \sqrt[3]{27 x}$ are inverse functions.

Step 1: Show that $f [g (x)] = x$ .

Substitute for g(x)

$f [g (x)] = f (\sqrt[3]{27 x})$

Substitute the expression in f(x)

$f (\sqrt[3]{27 x}) = \frac{{(\sqrt[3]{27 x})}^{3}}{27}$

Solve

$\frac{{(\sqrt[3]{27 x})}^{3}}{27} = \frac{27 x}{27} = x$

Step 2: Show that $g [f (x)] = x$ .

Substitute for f(x)

$g [f (x)] = g (\frac{x^{3}}{27})$

Substitute the expression in g(x)

$g (\frac{x^{3}}{27}) = \sqrt[3]{27 (\frac{x^{3}}{27})}$

Solve

$\sqrt[3]{27 (\frac{x^{3}}{27})} = \sqrt[3]{x^{3}} = x$

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.

$f (x) = \frac{x^{3}}{27} : (0, 0), (1, \frac{1}{27}), (2, \frac{8}{27}), (3, 1)$

$g (x) = \sqrt[3]{27 x} : (0, 0), (\frac{1}{27}, 1), (\frac{8}{27}, 2), (1, 3)$

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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