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151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
November 14
Things you should know:
f (x) increasing ⇔ f 0 (x) > 0
f (x) decreasing ⇔ f 0 (x) < 0
f (x) concave up ⇔ f 00 (x) > 0 ⇔ f 0 (x) is increasing
f (x) concave down ⇔ f 00 (x) < 0 ⇔f 0 (x) is decreasing
1. Here is the graph of f 00 (x). The domain is given and it is all real numbers. Where is f (x)
concave up? concave down? What are the inflection values of f (x).
The graph of f 0 (x)
E
A
B
C
D
This is the graph of f 00 (x). Our conclusions below are based on the fact that we are looking at
a picture of f 00 (x).
Since the function, f (x), is concave up when f 00 (x) is positive, then we look at the graph to see
where f 00 (x) is above the x-axis. (A, C), (C, D), (D, E).
Since the function, f (x), is concave down when f 00 (x) is negative, then we look at the graph to
see where f 00 (x) is below the x-axis. (−∞, A), (E, ∞).
The inflection values are x = A and x = E.
2. Now lets redo number 1; but this time let it be the graph of f 0 (x).
The graph of f 0 (x)
E
A
B
C
D
The function f (x) will be concave up when f 0 (x) is increasing: (−∞, A), (C, D)
The function f (x) will be concave down when f 0 (x) is decreasing: (B, C), (E, ∞).
The inflection values will be x = B, x = C, and x = D.
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151 WebCalc Fall 2002-copyright Joe Kahlig
3. Find the asymptotes for these functions. For part A look at the Asymptote handout to see the
shortcut.
x2 − 6x + 8
3x2 − 48
Horizontal asymptote: y = 13
Vertical asymptote: x = −4
(a) y =
(b) y = ln(x2 − 25)
if you take the limit at x → ∞ or x → −∞, you get an answer of ∞. Thus there are no
horizontal asymptotes.
To find the vertical asymptote, find the number that you would plug in to get ln 0. That
value of x will be a vertical asymptote.
Horizontal asymptote: none
Vertical asymptote: x = −5, x = 5
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(c) y = 3 + e−x
This function is never undefined, so no vertical asymptote. If you take the limit as x → ∞
or x → −∞, you get an answer of 3.
Horizontal asymptote: y = 3
Vertical asymptote: none
4. Find the intervals where the function is concave up and concave down, increasing and decreasing.
Find anything else that might be useful to graph the function.
3x + 4
−3x + 7
6(x − 6)
f (x) =
f 0 (x) =
f 00 (x) =
2
3
(x + 5)
(x + 5)
(x + 5)4
HA: y = 0
VA: x = −5
Setting the first derivative equal to zero and solving for x gives x = 73 . Note x = −5 is not a
critical value since it is not in the domain.
From the sign chart listed below, we can find the intervals of increasing and decreasing.
inc −5, 73
dec (−∞, −5),
7
3, ∞
−
y
7
3
There is a rel max at x =
There looks to be a rel min at x = −5.
However, that value is not a critical
value so it can not be a relative min.
−7
y
−
+
−5
−
7/3
0
5
+
−
−5
0
6
10
Taking the second derivative and setting it equal to zero gives a possible inflection values of
x = 6. Note x = −5 is not a possible inflection value since it is not in the domain.
From the sign chart listed above, we can find the intervals of concavity.
concave up: (6, ∞)
Concave down: (−∞, −5), (−5, 6)
The graph of the function is
5. Sketch a graph of f (x) using the supplied information.
(a) Assume that the function is defined and continuous for all real x.
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151 WebCalc Fall 2002-copyright Joe Kahlig
−5
6
y 0 undefined
y0 < 0
y0 = 0
y0 > 0
x1
x2
y0 < 0
x3
y 00 undefined
y 00 > 0
y 00 < 0
y 00 > 0
This is a rough sketch of the function. Nothing in the problem says where the function
should be placed on the graph, so there are many possible correct answers.
x1
x2
x3
(b) x-intercept of 0, vertical asymptote: x = −5 and x = 5
f 0 (−2) = f 0 (2) = f 0 (7) = 0, lim f (x) = 0, lim f (x) = 3
x→∞
x→−∞
f 0 (x) < 0 on (−∞, −5), (−5, −2) and (2, 5)
f 0 (x) > 0 on (−2, 2) and (5, ∞)
f 00 (x) > 0 on (−5, 0) and (7, 9)
f 00 (x) < 0 on (−∞, −5), (0, 5), (5, 7) and (9, ∞)
−2
−5
7
2
5
9