Inverse(^) functions
P1^
4
ExaMPlE 4.5 Find f−^1 (x) when f(x) = 2 x − 3 and the domain of f is x 4.
SOlUTION
Domain Range
Function: y = 2 x − 3 x 4 y 5
Inverse function: x = 2 y − 3 x 5 y 4
Rearranging the inverse function to make y the subject: y = x+ 23.
The full definition of the inverse function is therefore:
f−^1 (x) = x+ 23 for x 5.
You can see in figure 4.10 that the inverse function is the reflection of a restricted
part of the line y = 2 x − 3.
ExaMPlE 4.6 (i) Find f−^1 (x) when f(x) = x^2 + 2, x 0.
(ii) Find f(7) and f−^1 f(7). What do you notice?
SOlUTION
(i) Domain Range
Function: y = x^2 + 2 x 0 y 2
Inverse function: x = y^2 + 2 x 2 y 0
Rearranging the inverse function to make y its subject: y^2 = x − 2.
This gives y = ± x− 2 , but since you know the range of the inverse function
to be y 0 you can write:
y = + x− 2 or just y = x− 2.
y
O x
y = x
y = f(x)
y = f–1(x)
(4, 5)
(5, 4)
Figure 4.10