Concavity - 3.5
1. Concave up and concave down
Definition For a function f that is differentiable on an interval I, the graph of f is
a. If f is concave up on a, b , then the secant line passing through points x 1 , fŸx 1 and x 2 , fŸx 2 for any x 1 and x 2 in a, b are above the curve y fŸx between x 1 , fŸx 1 and x 2 , fŸx 2 .
b. If f is concave down on a, b , then the secant line passing through points x 1 , fŸx 1 and
x 2 , fŸx 2 for any x 1 and x 2 in a, b are below the curve y fŸx between x 1 , fŸx 1 and
x 2 , fŸx 2 .
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2
-3
-2
-1
1
x
2
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0
-3
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-1
6
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0
1
x
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a concave up function on "3, 3
a concave down function on "3, 3
Example The graph of f is given below. Determine graphically the interval on which f is
a. concave up;
b. concave up and decreasing.
15
10
a. f is concave up on 1. 8, 3. 3
b. f is concave up and decreasing on
5
1. 8, 2. 5
0
1
2
x
3
4
5
-5
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fŸx How can we determine algebraically where f is concave up and where f is concave down?
Theorem Suppose that f is differentiable on an interval I. The graph of f is
1
U
a. concave up on I, if f is increasing on I ; and
U
b. concave down on I, if f is decreasing on I.
UU
Or, suppose that f exists on I. The graph of f is
UU
a. concave up on I, if f Ÿx 0 for all x in I;
UU
b. concave down on I, if f Ÿx 0 for all x in I.
2. Inflection points:
Definition Suppose that f is continuous on the interval a, b . Let c be in a, b . Then the point
c, fŸc is called an inflection point of f if the graph of f changes concavity at the point
c, fŸc .
Note that the graph of f changes concavity at the point c, fŸc if
U
a. f changes from increasing to decreasing or from decreasing to increasing at c, fŸc or
b. f
UU
changes from positive to negative or from negative to positive at c, fŸc .
Example Let the graph of f UU Ÿx be given below. Find
a. the x "coordinate of each inflection point of f;
b. where the graph of f is concave up and is concave down.
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4
2
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1
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x
2
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-8
UU
f Ÿx UU
UU
a. f Ÿx 0 when x "2, x 0, x 1 and x 2. f does not change sign at x 0. So, the
x "coordinates of inflection points of f are x "2, x 1 and x 2.
UU
UU
b. f 0 for "2 x 0, 0 x 1, 2 x . and f 0 for ". x "2, 1 x 2. So,
the graph of f is concave up on "2, 0 : 0, 1 : 2, . and is concave down on
"., " 2 : 1, 2 .
Example Let the graph of f U Ÿx be given below. Find
a. the x "coordinate of each inflection point of f;
b. where the graph of f is concave up and is concave down.
2
15
10
5
0
1
2
x
3
4
5
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U
f Ÿx UU
f Ÿx 0 when x 0. 8, x 2. 5, x 4.
UU
U
f 0 Ÿf is increasing for 0 x 0. 8, 2. 5 x 4;
UU
U
f 0 f is decreasing for 0. 8 x 2. 5, 4 x 5.
a. So, x 0. 8, x 2. 5, x 4 are the x "coordinates of inflection points of f .
b. The graph of f is concave up on 0, 0. 8 : 2. 5, 4 and is concave down on 0. 8, 2. 5 : 4, 5 .
Example Let fŸx 2x 3 9x 2 " 24x " 10. Find
a. all possible inflection points of f;
b. where the graph of f is concave up and is concave down.
UU
Verify your answers by graphing both f and f .
Compute f
UU
:
U
UU
f Ÿx 6x 2 18x " 24, f Ÿx 12x 18 12Ÿ2x 3 UU
f Ÿx 0 if and only if x " 3 .
2
UU
Check signs of f over intervals: "., " 3 , " 3 , .
2
2
UU
UU
f Ÿ"2 12Ÿ"1 0, f Ÿ0 12Ÿ3 0
f " 3 79 .
2
2
UU
a. Since f changes sign at the point where x " 3 , " 3 , 79 is an inflection point of f.
2
2
2
3
b. The graph of f is concave up on " , . and is concave down on "., " 3 .
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2
3
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x
2
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- f , -.- f
UU
3. Second Derivative Test:
U
Theorem Suppose that f is continuos on the interval a, b and f Ÿc 0, for some c in a, b .
UU
a. If f Ÿc 0, then fŸc is a local maximum and
UU
b. if f Ÿc 0, then fŸc is a local minimum.
Example Let fŸx x 25
x . Find
a. the intervals of increase and decrease;
b. all local extrema;
c. the intervals of concavity;
d. all inflection points; and
e. sketch the graph of f based on the information in a.-d.
The domain of f : "., 0 , 0, .
U
Compute f and f
UU
:
2
U
f Ÿx 1 " 252 x "2 25 ,
x
x
U
Find critical numbers of f :
UU
f Ÿx "Ÿ"2 253 503
x
x
U
U
f Ÿx 0 « x 2 " 25 0, x o5; f Ÿx is not defined « x 0
Determine the sign change of f
U
over "., " 5 , "5, 0 , 0, 5 , 5, .
U
f Ÿo6 1 " 25 0, f Ÿo4 1 " 25 0
36
16
U
interval
U
f Ÿx Determine the sign change of f
4
UU
"., " 5
:
"5, 0
"
0, 5
"
5, .
UU
UU
f Ÿx p 0, f Ÿx changes sign as x 3 changes sign.
"., 0
interval
U
"
f Ÿx 0, .
State the results:
a. f is increasing on "., " 5 , 5, . and is decreasing on "5, 0 , 0, 5 .
b. By the first derivative test, fŸ"5 is a local maximum and fŸ5 is a local minimum.
c. f is concave up on 0, . and is concave down on "., 0 .
d. Though the graph of f changes concavity at x 0, x 0 is not an inflection point of f since it is not in
the domain of f, so there is no inflection point
e. Sketch the graph of f based on the information in a.-d
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0
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5 x
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Example Let fŸx Ÿx 2 1/5 4. Find
a. the intervals of increase and decrease;
b. all local extrema;
c. the intervals of concavity;
d. all inflection points; and
e. sketch the graph of f based on the information in a.-d.
The domain of f : "., .
U
Compute f and f
UU
:
UU
4
f Ÿx 1 " 4 Ÿx 2 "9/5 "
5
5
25Ÿx 2 9/5
1
f Ÿx 1 Ÿx 2 "4/5 ,
5
5Ÿx 2 4/5
U
Find critical numbers of f
U
:
U
f Ÿx p 0;
Determine the sign change of f
U
U
f Ÿx is not defined « x "2
over "., " 2 ,
"2, .
:
U
f 0 for all x, it does not change sign at x "2.
UU
Determine the sign change of f :
UU
UU
f Ÿx p 0, f Ÿx changes sign as Ÿx 2 9/5 changes sign.
interval
U
f Ÿx State the results:
5
"., " 2
"2, .
"
f is increasing on "., .
Since f is always increasing, there is no local extrema.
f is concave up on "., " 2 and is concave down on "2, . .
The graph of f changes concavity at x "2, "2, 4 is an inflection point of f .
Sketch the graph of f based on the information in a.-d
a.
b.
c.
d.
e.
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0 2
4 x6
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Example Sketch a graph of a function with the given properties:
fŸ0 2
U
U
f Ÿx 0, for all x; f Ÿ0 1
UU
f Ÿx 0 for x 0,
UU
UU
f Ÿx 0 for x 0, f Ÿ0 0
Example Sketch a graph of a function with the given properties:
fŸ0 0, fŸ"1 1, fŸ1 1
U
f Ÿx 0, for x "1 and 0 x 1,
U
f Ÿx 0 for " 1 x 0 and x 1;
UU
f Ÿx 0 for x 0 and x 0
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