03 8inversefuncs

# Inverse Functions

Recall that a function is a relation that is either one-to-one, or many-to-one.

• A function passes the vertical line test: any x value has no more than one y value.

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When we describe a function as one-to-one, this refers to

• each x value has only one y value
• each y value has only one x value
• A one-to-one function will pass the vertical line test and the equivalent horizontal line test.

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A one-to-one function that is continuous, can only be either:

• a strictly increasing function (increasing over its entire domain) (OR)
• a strictly decreasing function (decreasing over its entire domain)

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A function will have an inverse that is also a function if and only if it is one-to-one.

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## Notation

If a function, $f(x)$ is a one-to-one function,

• then its inverse function is denoted by $f^{-1}(x)$.

You are already familiar with this notation from trigonometry.

• the inverse of $\sin(x)$ is $\sin^{-1}(x)$

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### WARNING

Be careful not to confuse the notation for an inverse function $f^{-1}(x)$

with the notation for the reciprocal of a function: $\left( f(x) \right)^{-1} = \dfrac{1}{f(x)}$

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## Maximal Domain

The maximal domain (or implied domain) of a function is the largest domain for which its rule is defined.

If a function is given without a domain, it is assumed that the maximal domain is being used.

You may be asked to find the maximal domain for a function for which its inverse function is defined

• This usually implies that for the maximal domain, the inverse is not a function.
• You are being asked to state the largest restricted domain so that that section of the function is one-to-one
• ie the largest domain for which the inverse function, $f^{-1}(x)$ exists.

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#### Example 1

State the largest negative domain of $f(x) = (x + 3)^2$, so that $f^{-1}(x)$ exists.
Hence fully define the restricted function, $f(x)$ and its inverse function, $f^{-1}(x)$.

Solution

• Notice that the maximal (or implied) domain of $f(x)$ is $x \in R$.
• Notice that for the maximal domain, $f(x)$ is not one-to-one.

By observation of the graph, $f(x)$ is one-to-one for two equally large domains:

… … $x \in (-\infty, \; -3]$

… … $x \in [-3, \; +\infty)$

Hence, the largest negative domain for which the inverse function exists is: $x \in (–\infty, \; -3]$

and for this domain, the range of $f(x)$ is: $y \in [0, \; +\infty)$
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The inverse of f(x) can be found by swapping x and y in the rule:

… … Let $y=(x+3)^2$

… … reverse x and y

… … $x = (y+3)^2$

… … $(y+3)^2 = x$

… … $y+3=\pm \sqrt{x}$

… … $y=\pm \sqrt{x} -3$

but we want the range of $f^{-1}(x)$ to be the domain of $f(x): (–\infty, \; -3]$

so choose the negative half, hence:

… … $f^{-1}(x)=-\sqrt{x}-3$

Hence, fully defined, the function and its inverse function are:

… … $f:(-\infty,-3] \rightarrow R,\;\; f(x)=(x+3)^2$
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… … $f^{-1}: [0, +\infty) \rightarrow R, \;\; f^{-1}(x)=-\sqrt{x}-3$

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## Key Feature of Inverse Functions

A feature of inverse functions is that the composite of a function and its inverse will always give the variable (x) as the result.

… … $f \left( f^{-1}(x) \right) = x$

… … $f^{-1} \left( f(x) \right) = x$

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#### Example 2

… … a)$\sin \big(sin^{-1}(x) \big) = x$

… … b)$\sin^{-1} \big( sin(x) \big) = x$
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… … c)$\log_e \big(e^x \big) = x$

… … d)$e^{\log_e(x)} = x$
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… … e)$f(x)=(x+3)^2$

… … … .. $f^{-1}(x)=\sqrt{x}-3$

… … … .. $f \left( f^{-1}(x) \right) = \Big( \big( \sqrt{x}-3 \big) + 3 \Big)^2$

… … … … $\qquad \qquad \; = \big( \sqrt{x} \big)^2$

… … … … $\qquad \qquad \; = x$
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… … f)$f^{-1}(x) = \sqrt{x}-3$

… … … .. $f^{-1} \big( f(x) \big) = \sqrt{(x+3)^2} - 3$

… … … … $\qquad \qquad \; = (x + 3) - 3$

… … … … $\qquad \qquad \; = x$

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