Mathematics
Reciprocal Functions
Science and Mathematics
Education Research Group
Supported by UBC Teaching and Learning Enhancement Fund 2012-2014
Department of
Curr iculum and Pedagogy
F A C U L T Y O F E D U C A T I O N a place of mind
Reciprocal Functions
This question set deals with functions in the form:
Given the function f(x), we will analyze the shape of the graph of g(x).
Notice the difference between the reciprocal of a function, and the
inverse of a function:
The reciprocal and the inverse of a function are not the same.
1)(
)(
1)(
xf
xfxg
1)()(
xfxg )()( 1 xfxg
Reciprocal: Inverse:
)1
,(),(b
aba ),(),( abba
Reciprocal Functions I
Consider the value of (the reciprocal of p) , where .
Which of the following correctly describes the approximate value of
for varying values of ?
A.
p
10p
p
1
p
p is very large
(for example, p = 1000)
p is very small
(for example, p = 0.0001)
is large is large
is close to 0 is large
is large is close to 0
is close to 0 is close to 0
B.
C.
D.
E.
p
1
p
1
p
10
1
p
p
1
p
1
p
1
p
1
p
1
p
1
Solution
Answer: B
Justification: Fractions with large denominators are smaller than
fractions with small denominators (for equal positive numerators).
Consider when p = 1000. Its reciprocal is a very small positive
number:
As p becomes larger, its reciprocal becomes smaller.
Now consider when p = 0.001. Its reciprocal is a number much
greater than 1:
As p becomes smaller, its reciprocal becomes larger.
001.01000
11
p
1000
1000
1
1
001.0
11
p
Reciprocal Functions II
Consider the function .
When , what are the possible values for ?
1)(E.
0)(D.
1)(0C.
1)(B.
0)(A.
xg
xg
xg
xg
xg
1)()(
xfxg
1)(0 xf )(xg
Press for hint
)(
1)(
)()(1
xfxg
xfxg
x
x
aa
1
Exponent laws:
Solution
Answer: B
Justification: Try choosing a value for , such as .
In this case,
Since , which is greater than 1, we can rule out C, D, and E.
We must check if it is possible for to be between 0 and 1 in
order to choose between answers A and B.
Notice that is only between 0 and 1 when the numerator
is smaller than the denominator. This is not possible, because the
denominator is never greater than 1.
Therefore, when .
)(xf2
1)( xf
2
2
1
1
)(
1)()(
1
xfxfxg
2)( xg
)(xg
)(1)(
xfxg
)1)(0( xf
1)( xg 1)(0 xf
Reciprocal Functions III
Consider a function and its reciprocal function .
Suppose that at , . What are all the possible
values for ?
1,0)(E.
1)(D.
1)(C.
0)(B.
1)(A.
pf
pf
pf
pf
pf
)(xf 1)()(
xfxg
)()( pgpf
)( pf
px
Solution
Answer: D
Justification: This question asks to find the values of f(p) where
. We must find a value that is equal to its reciprocal.
We can get the answer directly by solving the above equation for f(p):
If we tried to plug in , we get which is undefined.
Conclusion: Points where are also points on the reciprocal
of f. Points where form vertical asymptotes on the reciprocal
function.
)(
1)(
pfpf
1
11,
1
11
0)( pf 01)( pg
1)( xf
0)( xf
1)(1)()(
1)(
2 pfpf
pfpf Notice:
Reciprocal Functions IV
Let .
Which of the following correctly describes when and
when ?
A.
)(
1)(
xfxg
B.
C.
D.
E.
)(xg 0)( xf
0)( xf
0)( xf 0)( xf
0)( xg 0)( xg
0)( xg 0)( xg
0)( xg 0)( xg
0)( xg 0)( xg
1)(0 xg 0)(1 xg
Solution
Answer: C
Justification: When , must also be greater than
zero. The numerator is positive, and the denominator is positive. A
positive number divided by a positive number is positive.
When , must also be less than zero. The numerator
is positive but the denominator is negative. A positive number
divided by a negative number is negative.
Conclusion: If is above the x-axis, is also above the x-
axis. When is below the x-axis, is also below the x-axis.
0)( xf
0)( xf
)(xf )(xg
)(xf )(xg
)(1
xf
)(1
xf
Reciprocal Functions V
If the point , where , lies on the graph of , what point
lies on the graph of ?
ba
ba
ba
ab
ba
1,
1E.
,1
D.
1,C.
,B.
,A.
),( ba f
)(
1)(
xfxg
0b
Notice that this question has the restriction
that b ≠ 0. What would happen to the point
on the reciprocal function if b = 0?
Solution
Answer: C
Justification: The reciprocal function takes the
reciprocal of the y-value for each point in . Consider when x = a:
Notice that x-values are not changed. The point transforms
into the point .
Do not get confused with the inverse of a function ,
which interchanges x and y-values. In general:
1)()(
xfxg
f
)()( 1 xfxg
11 )()( xfxf
bag
afag
1)(
)(
1)(
when x = a
since f(a) = b
),()(, baafa
baaga
1,)(,
),( ba
ba
1,
Summary
)(xf 1)(
xf
1)( xf
1)( xf
0)(1 xf
0)( xf
1)(0 xf
1)( xf
1)( xf
1)(1
xf
1)(1
xf
1)(01
xf
1)(1
xf
asymptote vertical
1)(1
xf
0)(11
xf
Strategy for Graphing I
Consider the function and its reciprocal function .
The following example outlines the steps to graph :
1. Identify the points where or . These points exist
on the reciprocal function. (See question 3)
1)( xxf1
1)(
xxg
1)( xf
g
1)( xf
1)( xxf
1y
1y
1
1)(
xxg
)1,0(
)1,2(
Strategy for Graphing II
2. Identify the points where . Draw a dotted vertical line
through these points to show the locations of the vertical
asymptotes. (Recall taking the reciprocal of 0 gives an undefined
value.)
0)( xf
0)1( f
1)( xxf
1
1)(
xxg
1x
1
1)(
xxg
1x
Strategy for Graphing III
3. Consider when . Draw a curve from the top of the
asymptote line to the point where . When ,
draw the curve from the bottom of the asymptote to .
1)(0 xf
1)( xg
1)( xxf
1)( xg
0)(1 xf
The reciprocal of
small values become
large values.
Strategy for Graphing IV
4. Consider where . From the point , draw a curve
that approaches the x-axis from above. When , the curve
approaches the x-axis from below.
1)( xf
1
1)(
xxg
1x
1)( xxf
1)( xf
1)( xf
The reciprocal of
large values become
small values.
23)( xxf
Reciprocal Functions VI
The graph of is shown
below. What is the correct
graph of its reciprocal
function, ?
f
g
A. B.
C. D.
Solution
Answer: D
Justification: Since , its reciprocal must also pass through
the point (1,1). This eliminates answers B and C.
1)1( f
Additionally, f is positive for and
negative for .
This must also be true for the reciprocal
function, so answer A must be eliminated
because it is always positive. The only
remaining answer is D.
What other features can be used to
identify the reciprocal function? What can
we say about points (1,1) and ?
32x
32x
)1,3
1(
Reciprocal Functions VII
The graph of is shown
below. What is the correct
graph of its reciprocal
function, ?
f
g
A. B.
C. D. f
Solution
Answer: A
Justification: When , . The reciprocal function also
has this horizontal line since the reciprocal of 1 is still 1. This
eliminates answers B and D.
2x
Since f(0) = 0, g(x) must have an
asymptote at x = 0. This only leaves
function A.
1)( xf
Reciprocal Functions VIII
The graph of is shown
below. What is the correct
graph of its reciprocal
function, ?
f
g
A. B.
C. D. f
Solution
Answer: C
Justification: Consider the point (0, 2) on . This point will
appear as on . Only function C passes through this point. )2
1,0( g
f
What other features can be used to
identify the reciprocal function?
Reciprocal Functions IX
The graph of is shown
below. What is the correct
graph of its reciprocal
function, ?
f
g
A. B.
C. D.
f
Solution
Answer: A
Justification: The reciprocal function must be positive when the
original function is positive, and negative when the original function
is negative. This eliminates answers B and D.
The reciprocal function must have
asymptotes at because the
original function has zeroes at these
values of x. Only graph A satisfies
both of these conditions.
2x
25.0)( 2 xxf
25.0
1)(
2
xxg
Solution
Answer: D
Justification: Find where the original function
crosses the lines :
12)( xxf
0,1y
1
121
x
x
2
1
120
x
x
0
121
x
x
passes through
the point (1, -1).
1)(
xf passes through
the point (0, 1).
has an
asymptote at x = 0.5.
Only graph D satisfies these 3 conditions.
This question can also be solved by first graphing ,
then determining the reciprocal graph.
12)( xxf
1)(
xf 1
)(
xf
Reciprocal Functions XI
The graph shown to the
right is in the form:
What are the values of m
and b?
bmx
xf
1
)(1
1,5.0E.
1,2D.
1,5.0C.
1,2B.
1,1A.
bm
bm
bm
bm
bm
Solution
Answer: C
Justification: The function f is in
the form . We need to
find a pair of points that lie on f to
determine its equation. One good
choice is (-4, 1) and (0, -1), since
these points lie on both f and f-1.
Another choice for a point is (-2, 0),
since there is an asymptote at x = -2.
Using any of the 3 points mentioned
above, find the slope of the line:
The y-intercept is at (0, -1), so
bmxxf )(
5.0)4(0
1)1(
m
1b
)1,4(
)1,0(
Points that lie on the
green lines belong to
both f and its reciprocal
)0,2(
Reciprocal Functions XII
What is the graph of the
cotangent function?
)tan(
1)cot()(
1
xxxf
A. B.
C. D.
)tan()( xxf
0 5.05.0
0 5.05.0 0 5.05.0
0 5.05.0 0 5.05.0
Solution
Answer: B
Justification: The points that lie
on both the tangent and cotangent
function are and .
The tangent function also has
asymptotes at , where k
is an integer. At these values of x,
the cotangent function must return
0. The cotangent function has
asymptotes at x = kπ.
1,4
1,4
kx 2
0 5.05.0)tan()( xxf
)tan(
1)cot()(
1
xxxf