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In earlier grades, you studied various types of functions and understood the effect of various parameters in the general equation. In this section, we will consider inverse functions .
An inverse function is a function which "does the reverse" of a given function. More formally, if $f$ is a function with domain $X$ , then ${f}^{-1}$ is its inverse function if and only if for every $x\in X$ we have:
A simple way to think about this is that a function, say $y=f\left(x\right)$ , gives you a $y$ -value if you substitute an $x$ -value into $f\left(x\right)$ . The inverse function tells you tells you which $x$ -value was used to get a particular $y$ -value when you substitue the $y$ -value into ${f}^{-1}\left(x\right)$ . There are some things which can complicate this for example, think about a $sin$ function, there are many $x$ -values that give you a peak as the function oscillates. This means that the inverse of a $sin$ function would be tricky to define because if you substitute the peak $y$ -value into it you won't know which of the $x$ -values was used to get the peak.
This works both ways, if we don't have any complications like in the case of the $sin$ function, so we can write:
For example, if the function $x\to 3x+2$ is given, then its inverse function is $x\to {\displaystyle \frac{(x-2)}{3}}$ . This is usually written as:
The superscript "-1" is not an exponent.
If a function $f$ has an inverse then $f$ is said to be invertible.
If $f$ is a real-valued function, then for $f$ to have a valid inverse, it must pass the horizontal line test , that is a horizontal line $y=k$ placed anywhere on the graph of $f$ must pass through $f$ exactly once for all real $k$ .
It is possible to work around this condition, by defining a “multi-valued“ function as an inverse.
If one represents the function $f$ graphically in a $xy$ -coordinate system, the inverse function of the equation of a straight line, ${f}^{-1}$ , is the reflection of the graph of $f$ across the line $y=x$ .
Algebraically, one computes the inverse function of $f$ by solving the equation
for $x$ , and then exchanging $y$ and $x$ to get
The inverse function of $y=ax+q$ is determined by solving for $x$ as:
Therefore the inverse of $y=ax+q$ is $y=\frac{1}{a}x-\frac{q}{a}$ .
The inverse function of a straight line is also a straight line, except for the case where the straight line is a perfectly horizontal line, in which case the inverse is undefined.
For example, the straight line equation given by $y=2x-3$ has as inverse the function, $y=\frac{1}{2}x+\frac{3}{2}$ . The graphs of these functions are shown in [link] . It can be seen that the two graphs are reflections of each other across the line $y=x$ .
We have seen that the domain of a function of the form $y=ax+q$ is $\{x:x\in \mathbb{R}\}$ and the range is $\{y:y\in \mathbb{R}\}$ . Since the inverse function of a straight line is also a straight line, the inverse function will have the same domain and range as the original function.
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