1.9
Inverses
JMerrill, 2010
Inverses

Graph the function
f(x) = 2x on the
graph and record
some ordered pairs.
x
0
1
f(x)
1
2
2
3
4
8
Function

Is this a function?


Yes, no x’s are
repeated. (It
passes the vertical
line test.)
3. Domain?
Range?
 ,  
 0,  
One-to-One Functions

A function is oneto-one (one x for
one y) and has an
inverse that is a
function if it passes
the horizontal line
test
Graphing Inverses
To graph the inverse of a function, simply
switch the x’s and y’s
f(x) =
x
f(x)
0
1
1
2
2
4
3
8
f-1(x) =
Inverse
notation
x
f(x)
1
0
2
1
4
2
8
3
Now graph

f(x)

f-1(x)
Putting Both on the Same Graph
Graphically, you can
tell that the
functions are inverses
of each other because
they are symmetric
about the line y = x
How are the Domain and Range of
f(x) and f -1(x) related?
The domain of the original function is the
same as the range of the new function
and vice versa.
f(x) =
x
f(x)
x
f(x)
0
1
1
0
1
2
2
1
2
4
4
2
3
8
8
3
f -1(x) =
Proving Functions are Inverses


The only way to prove that functions
are inverses of each other is to do a
composition of functions (f(g(x)) and
g(f(x)) and have both answers equal to
x
Both answers must be X, not just the
same answer.
Example



Given f(x) = 3x+1 and g(x) = x  1
3
Prove they are inverses of each other
f(g(x))
g(f(x))
x 1
f

3


x 1
3
 1
 3 
x 1 1  x
g(3x  1)
(3x  1)  1
3
3x
x
3
You Do


Given f(x) = 2x2 and g(x) = x
2
Prove they are inverses of each other
f(g(x)) 
 x
f

 2
2
 x
2
 
 2
x

2   x
2
g(f(x) 
g(2x2 ) 
2x

2
2
x2  x
Writing the Inverse Function


Given f(x) = 3x + 5, find f-1(x)
Rewrite the function in terms of y


Switch the x’s and y’s



y = 3x+5
x = 3y+5
x5
y
3
Solve for y:
Rename the function:
f1 (x) 
x5
3
You Do

Given f(x) = x2 + 5, find f-1(x)
y  x2  5
x  y 5
2
x5  y
2
x5  y
Note: We had to restrict the
domain in the original
function to (0,∞) since the
function does not pass the
horizontal line test.
Therefore, when we take the
square root of both sides, we
only have the positive value.