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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


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.

Strategy for Graphing V

The final graph of is shown below. 1

1)(

xxg

1

1)(

xxg

1

1)(

xxg

1)( xxf

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




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




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




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

Reciprocal Functions X

What is the graph of the

reciprocal function

?12

1)(

1

xxf

A. B.

C. D.

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

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