1-3 Continuity, End Behavior, and Limits
Use logical reasoning to determine the end behavior or limit of the function as x approaches infinity.
Explain your reasoning.
33. SOLUTION: Graph q(x). Explore what happens as x → ∞.
x
–1000
–100
–10
–1
0
1
10
100
1000
g(x)
0.024
0.24
2.4
24
-–24
–2.4
–0.24
–0.024
From the table you can see that as x approaches negative infinity g(x) decreases to 0. as x approaches positive
infinity g(x) decreases to 0. Algebraically, as x increases, the denominator will increase. This will make the fraction get closer and closer to zero.
The negative sign indicates that q(x) will approach 0 from the negative side.
35. p (x) =
SOLUTION: Graph p (x). eSolutions Manual - Powered by Cognero
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Algebraically, as x increases, the denominator will increase. This will make the fraction get closer and closer to zero.
The negative sign indicates that q(x) will approach 0 from the negative side.
1-3 Continuity, End Behavior, and Limits
35. p (x) =
SOLUTION: Graph p (x). Explore what happens as x → ∞.
x
–1000
–100
–10
–1
0
1
10
100
1000
p (x)
0.997
0.971
0.75
0
–0.5
–2
1.375
1.031
1.003
As x increases, the numerator and denominator will increase. For larger values of x, the "+1" and "–2" will have less
of an affect on the values. For example,
So, as x increases, the value of the fraction approaches
.
37. SOLUTION: Graph c(x).
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1-3 So,
Continuity,
End Behavior, and Limits
as x increases, the value of the fraction approaches
.
37. SOLUTION: Graph c(x).
Explore what happens as x → ∞.
x
–1000
–100
–10
–1
0
1
10
100
1000
c(x)
–0.005
–0.05
–0.49
–2.5
0
1.25
0.49
0.05
0.005
As x increases, the numerator and denominator will increase. For larger values of x, the "+1" and will have less of an affect on the values. For example,
So, as x increases, the value of the fraction approaches 0. Thus, as x
, c(x) will approach 0.
39. h(x) = 2x5 + 7x3 + 5
SOLUTION: Graph h(x).
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1-3 Continuity, End Behavior, and Limits
So, as x increases, the value of the fraction approaches 0. Thus, as x
, c(x) will approach 0.
39. h(x) = 2x5 + 7x3 + 5
SOLUTION: Graph h(x).
Explore what happens as x → ∞.
x
–10
–1
0
1
10
h(x)
–206,995
–4
5
14
207,005
As x increases, the h(x) will increase without bound.
Thus as x
, h(x) will approach ∞.
GRAPHING CALCULATOR Graph each function and determine whether it is continuous. If
discontinuous, identify and describe any points of discontinuity. Then describe its end behavior and
locate any zeros.
45. SOLUTION: Using the TRACE function we can see that the graph has discontinuities near x = −1, x = 2, and x = 3.
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1-3 Continuity, End Behavior, and Limits
Using the TRACE function we can see that the graph has discontinuities near x = −1, x = 2, and x = 3.
Use the TABLE function to confirm the points.
Since the graph approaches positive or negative infinity as it approaches x = −1, x = 2, and x = 3, these points are
infinite discontinuities. The graph appears to be approaching zero at both positive and negative infinity. Construct a table using extreme
values for x to confirm this.
From the table we can see that f (x) approaches zero for both extremely positive and negative values for x. eSolutions
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Press 2nd, CALC and select the zero option to determine the zeroes of the graph. We can see that this function
has a zero at x = 0.
1-3 Continuity, End Behavior, and Limits
From the table we can see that f (x) approaches zero for both extremely positive and negative values for x. Press 2nd, CALC and select the zero option to determine the zeroes of the graph. We can see that this function
has a zero at x = 0.
The graph has infinite discontinuities at x = −1, x = 2, and x = 3.
There is a zero at x
= 0.
47. SOLUTION: Using the TRACE function we can see that the graph has discontinuities near x = −6 and x = 3.
Use the TABLE function to confirm this.
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1-3 Continuity, End Behavior, and Limits
Use the TABLE function to confirm this.
Since the graph approaches positive or negative infinity as it approaches x = −6 and x = 3, these points are infinite
discontinuities. The graph appears to be approaching 4 at both positive and negative infinity. Construct a table using extreme values
for x to confirm this.
From the table we can see that f (x) approaches 4 for both extremely positive and negative values for x. Press 2nd, CALC and select the zero option to determine the zeroes of the graph. We can see that this function
has zeros at x = −3 and x =
.
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The graph has infinite discontinuities at x = 3 and x = −6.
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There are zeros at x = −3
1-3 Continuity, End Behavior, and Limits
The graph has infinite discontinuities at x = 3 and x = −6.
and x =
There are zeros at x = −3
.
49. SOLUTION: Using the TRACE function we can see that the graph has discontinuities near x = −4 and x = 3.
Use the TABLE function to confirm this.
Since the graph approaches positive or negative infinity as it approaches x = −4 and x = 3, these points are infinite
discontinuities. The graph appears to be approaching negative infinity as x approaches negative infinity, and positive infinity as x
approaches positive infinity. Construct a table using extreme values for x to confirm this.
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Since the graph approaches positive or negative infinity as it approaches x = −4 and x = 3, these points are infinite
discontinuities. 1-3 The
Continuity,
End Behavior, and Limits
graph appears to be approaching negative infinity as x approaches negative infinity, and positive infinity as x
approaches positive infinity. Construct a table using extreme values for x to confirm this.
From the table we can see that f (x) becomes largely negative as x becomes more negative. f (x) becomes largely
positive as x becomes more positive. Press 2nd, CALC and select the zero option to determine the zeroes of the graph. We can see that this function
has a zero at x = −5, x = 4, and x = 6.
The graph has infinite discontinuities at x = −4 and x = 3.
There are zeros at x =
−5, x = 4, and x = 6.
GRAPHING CALCULATOR Graph each function, and describe its end behavior. Support the
conjecture numerically, and provide an effective viewing window for each graph.
51. f (x) = −x4 + 12x3 + 4x2 – 4
SOLUTION: 4
3
2
Enter Y1 = –x + 12x + 4x – 4 on a graphing calculator.
Sample Method 1: Start by graphing Y1 using the standard viewing window. Use the Zoom Out feature or choose
larger values for the window dimensions to obtain a more complete view of the graph of the function. Continue this
process until you have obtained an effective view of the end behavior and the important points in between for the
function.
Sample Method 2: Use the Table feature and scroll in both directions from zero to determine an interval for x that
will contain all the changes in sign and direction for the y-values. Determine the maximum and minimum y-values
over this interval for x to determine the interval to use for y. Enter the window settings and graph.
For this function, there are changes in sign between the x-values 0 and 1 and between 12 and 13 and a change in
direction occurs when y is about 2500. So, an effective viewing window can be obtained using an interval of [–20,
20] for x and [–2500, 2500] for y. Choose a reasonable scale for each interval.
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The graph has infinite discontinuities at x = −4 and x = 3.
There are zeros at x =
−5, x = 4, and x = 6.
1-3 Continuity, End Behavior, and Limits
GRAPHING CALCULATOR Graph each function, and describe its end behavior. Support the
conjecture numerically, and provide an effective viewing window for each graph.
51. f (x) = −x4 + 12x3 + 4x2 – 4
SOLUTION: 4
3
2
Enter Y1 = –x + 12x + 4x – 4 on a graphing calculator.
Sample Method 1: Start by graphing Y1 using the standard viewing window. Use the Zoom Out feature or choose
larger values for the window dimensions to obtain a more complete view of the graph of the function. Continue this
process until you have obtained an effective view of the end behavior and the important points in between for the
function.
Sample Method 2: Use the Table feature and scroll in both directions from zero to determine an interval for x that
will contain all the changes in sign and direction for the y-values. Determine the maximum and minimum y-values
over this interval for x to determine the interval to use for y. Enter the window settings and graph.
For this function, there are changes in sign between the x-values 0 and 1 and between 12 and 13 and a change in
direction occurs when y is about 2500. So, an effective viewing window can be obtained using an interval of [–20,
20] for x and [–2500, 2500] for y. Choose a reasonable scale for each interval.
[–20, 20] scl:2 by [–2500, 2500] scl:500
end behavior:
= −
,
= −
To support this conjecture numerically, make a table showing the values of f (x) using very small and very large
powers of 10 for x.
So, the table supports the conjecture that as x approaches ∞, f (x) approaches –∞ and as x approaches –∞, f (x)
approaches –∞.
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SOLUTION: Page 10
So, the table supports the conjecture that as x approaches ∞, f (x) approaches –∞ and as x approaches –∞, f (x)
1-3 approaches
Continuity,
–∞. End Behavior, and Limits
53. SOLUTION: Enter Y1 =
on a graphing calculator.
Sample Method 1: Start by graphing Y1 using the standard viewing window. Use the Zoom Out feature or choose
larger values for the window dimensions to obtain a more complete view of the graph of the function. Continue this
process until you have obtained an effective view of the end behavior and the important points in between for the
function.
Sample Method 2: Use the Table feature and scroll in both directions from zero to determine an interval for x that
will contain all the changes in sign and direction for the y-values. Determine the maximum and minimum y-values
over this interval for x to determine the interval to use for y. Enter the window settings and graph.
For this function, there are changes in sign between the x-values –16 and –14 and between –1 and 1. There is a
vertical asymptote at x = –15 and the ends of the graph appear to approach a horizontal line. So, an effective
viewing window can be obtained using an interval of [–80, 80] for x and [–80, 80] for y. Choose a reasonable scale
for each interval.
end behavior:
= 16, = 16
To support this conjecture numerically, make a table showing the values of f (x) using very small and very large
powers of 10 for x.
So, the table supports the conjecture that as x approaches ∞, f (x) approaches 16 and as x approaches –∞, f (x)
approaches 16.
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