When finding the equation of an inverse function, assuming the inverse function exists, there's a nice simple process we can adopt.
Since we know, and can see visually, that the domain of the original function becomes the range of the inverse function (and the range of the original function becomes the domain of the inverse function), an elegant way to find the inverse function is to first swap the location of $x$xand $y$y in the original function. Effectively by doing this we're noting that we're swapping the domain and range for the inverse function. We'll then rearrange the function so that we have $y$y in terms of $x$x.
Worked Examples
example 1
Find the inverse function of $f\left(x\right)=\frac{1}{x+1}$f(x)=1x+1
Think: Note that this is the equation of a reciprocal function and is one-to-one, hence there will be an inverse function over a natural domain. Although there is no $y$y in the original function, we know that $f\left(x\right)=y$f(x)=y
Do:
$x=\frac{1}{y+1}$x=1y+1
$y+1=\frac{1}{x}$y+1=1x
$y=\frac{1}{x}-1$y=1x−1
With the power of CAS, we can also determine the inverse with a few commands, (but you will be able to answer the questions that follow by hand if you don't have a CAS).
example 2
Find the inverse function of $y=\ln\left(x\right)$y=ln(x)
Think: Remember that the natural logarithm has a base of $e$e.
Do:
$x=\ln\left(y\right)$x=ln(y)
By converting the logarithmic expression to an exponential expression:
$e^x=y$ex=y
Via CAS:
More Examples
Question 1
Consider the function $f\left(x\right)=x+1$f(x)=x+1.
Graph the function on the set of axes below:
Loading Graph...Is the function $f\left(x\right)=x+1$f(x)=x+1 a one-to-one function?
No
AYes
BIf a function is one-to-one, an inverse function exists. Find the inverse function $f^{-1}\left(x\right)$f−1(x), writing each line of working as an equation:
Question 2
Consider the function $f\left(x\right)=x^2-8x+11$f(x)=x2−8x+11.
Graph the function on the axes below:
Loading Graph...Is the function $f\left(x\right)=x^2-8x+11$f(x)=x2−8x+11 a one-to-one function?
No
AYes
BAlthough the function is not one-to-one, if we restrict the domain we can find a portion of the function that is one-to-one.
Which of the following domains are appropriate restrictions for $f\left(x\right)$f(x) to be a one-to-one function?
Select all that apply.
$\left[1,\infty\right)$[1,∞)
A$\left[4,\infty\right)$[4,∞)
B$\left(-\infty,8\right]$(−∞,8]
C$\left(-\infty,4\right]$(−∞,4]
DFind the inverse function $f^{-1}\left(x\right)$f−1(x) for $f\left(x\right)$f(x) on the restricted domain $\left[4,\infty\right)$[4,∞), by replacing $x$x with $y$y and $f\left(x\right)$f(x) with $x$x solving for $y$y.
Question 3
We want to determine whether $f\left(x\right)=3x+12$f(x)=3x+12 and $g\left(x\right)=\frac{1}{3}x-4$g(x)=13x−4 are inverse functions.
Find $f\left(g(x)\right)$f(g(x)).
Find $g\left(f(x)\right)$g(f(x)).
Are $f\left(x\right)$f(x) and $g\left(x\right)$g(x) inverses?
Yes
ANo
B