Remember the inverse of a function, written as $f^{-1}$f−1, swaps the order of the inputs and outputs of the original function $f$f.
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| A function $f$f and its inverse $f^{-1}$f−1. |
Because of this, we can view the outputs of the original function $f$f as the inputs for the function $f^{-1}$f−1. In the same way, we can view the inputs of the original function $f$f as the outputs for the function $f^{-1}$f−1.
For this reason, the function $f^{-1}$f−1 reverses the domain and range of the function $f$f.
The domain of the inverse function $f^{-1}$f−1 is equal to the range of the function $f$f.
The range of the inverse function $f^{-1}$f−1 is equal to the domain of the function $f$f.
Worked example
Find the inverse function of $f\left(x\right)=-\sqrt{x-3}$f(x)=−√x−3 defined over $\left[3,\infty\right)$[3,∞) and state its domain and range.
Think: To find the inverse function we would like to express the argument of $x$x in terms of $f\left(x\right)$f(x). Then we can determine the domain and range of $f^{-1}$f−1 by considering the domain and range of $f$f.
Do:
| $f\left(x\right)$f(x) | $=$= | $-\sqrt{x-3}$−√x−3 | (Writing down the equation for $f\left(x\right)$f(x)) |
| $f\left(x\right)^2$f(x)2 | $=$= | $x-3$x−3 | (Squaring both sides) |
| $f\left(x\right)^2+3$f(x)2+3 | $=$= | $x$x | (Making $x$x the subject) |
| $x$x | $=$= | $f\left(x\right)^2+3$f(x)2+3 | (Swapping the two sides) |
| $f^{-1}(x)$f−1(x) | $=$= | $x^2+3$x2+3 | (Rewriting the equation using $f^{-1}(x)$f−1(x)) |
The equation of the inverse function is $f^{-1}(x)=x^2+3$f−1(x)=x2+3.
The range of $f^{-1}$f−1 is equal to $\left[3,\infty\right)$[3,∞) since this is the domain of $f$f.
The domain of $f^{-1}$f−1 is equal to the range of $f$f. To find the range of $f$f we observe that when $x=3$x=3 the value of $f\left(x\right)$f(x) is $3$3. As $x$x tends to infinity, $f\left(x\right)$f(x) tends to negative infinity and so the domain of $f^{-1}$f−1 is $\left(-\infty,0\right]$(−∞,0].
Practice questions
question 1
The graph of $f$f is represented by the line segment joining $\left(-4,5\right)$(−4,5) and $\left(9,-4\right)$(9,−4) while $f^{-1}$f−1 is represented by the line segment joining $\left(5,-4\right)$(5,−4) and $\left(-4,9\right)$(−4,9).
State the domain of $f$f in interval notation.
Domain: $\editable{}$
State the range of $f$f in interval notation.
Range: $\editable{}$
State the domain of $f^{-1}$f−1 in interval notation.
Domain: $\editable{}$
State the range of $f^{-1}$f−1 in interval notation.
Range: $\editable{}$
Select the correct statement.
The range of $f$f is the same as the range of $f^{-1}$f−1.
AThe range of $f$f is the same as the domain of $f^{-1}$f−1.
B
question 2
Consider the function given by $f\left(x\right)=x+6$f(x)=x+6 defined over the interval $\left[0,\infty\right)$[0,∞).
Plot the function $f\left(x\right)=x+6$f(x)=x+6 over its domain.
Loading Graph...Find the inverse $f^{-1}$f−1.
State the domain and range of $f^{-1}$f−1 in interval notation.
Domain: $\editable{}$
Range: $\editable{}$
Plot the function $f^{-1}$f−1 over its domain.
Loading Graph...
question 3
Consider the many-to-one function $f\left(x\right)=x^2$f(x)=x2 defined for all real values of $x$x.
State the range of $f$f in interval notation.
Range: $\editable{}$
What is the smallest value of $a$a on the domain $\left[a,\infty\right)$[a,∞) such that function $f$f is invertible?
State the domain and range of the inverse of $f$f as defined in part (b) in interval notation.
Domain: $\editable{}$
Range: $\editable{}$