Inflection Points Practice Problems Solutions
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University of Maryland Global Campus (UMGC) *
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Course
140
Subject
Mathematics
Date
Apr 3, 2024
Type
docx
Pages
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Uploaded by ConstableKouprey17758
Inflection Points Practice Problems
At which point(s) does the graph to the right have
points of inflection?
Points A, C, and D
Let f
(
x
)
=
x
3
−
6
x
2
+
12
x
. Find the intervals where f
is
concave up and the intervals where f
is concave down.
f
'
(
x
)
=
3
x
2
−
12
x
+
12
f
''
(
x
)
=
6
x
−
12
f
''
(
x
)
=
0
6
x
−
12
=
0
6
x
=
12
x
=
2
Constructing a line test:
Concave up: (
2
,∞
)
Concave down: (−
∞,
2
)
Inflection Points Practice Problems
Consider g
(
x
)
=
2
x
3
−
3
x
2
−
12
x
+
5
. Discuss the concavity of the function. g
'
(
x
)
=
6
x
2
−
6
x
−
12
g
' '
(
x
)
=
12
x
−
6
g
' '
(
x
)
=
0
12
x
−
6
=
0
12
x
=
6
x
=
1
2
Constructing a line test:
Concave up: (
1
2
,∞
)
Concave down: (
−
∞,
1
2
)
g' '
(
x
)
Inflection Points Practice Problems
Find the point(s) of inflection of y
=
x
3
(
x
−
4
)
.
I personally would just distribute the x
3
through – if you did product rule, there is nothing wrong with that. y
=
x
4
−
4
x
3
y
'
=
4
x
3
−
12
x
2
y
' '
=
12
x
2
−
24
x
y
' '
=
0
12
x
2
−
24
x
=
0
12
x
(
x
−
2
)
=
0
x
=
0
and x
=
2
Careful! These are only candidates
for inflection points. In order to be an inflection point, the second derivative must change signs at those x
-values. Constructing a line test:
We know points of inflection will occur when the second derivative changes signs. We can see the second derivative changes signs at x
=
0
and x
=
2
.
Inflection points:
(
0
, y
(
0
)
)
→
(
0,0
)
(
2
, y
(
2
)
)
→
(
2
,
−
16
)
For what values(s) of x
does y
=−
2
x
4
−
8
x
3
+
180
x
2
have a point of inflection?
f
'
(
x
)
=−
8
x
3
−
24
x
2
+
360
x
f
''
(
x
)
=−
24
x
2
−
48
x
+
360
f
''
(
x
)
=
0
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Inflection Points Practice Problems
−
24
(
x
2
+
2
x
−
15
)
=
0
−
24
(
x
+
5
) (
x
−
3
)
=
0
x
=−
5
and x
=
3
Careful! These are only candidates
for inflection points. In order to be an inflection point, the second derivative must change signs at those x
-values. Constructing a line test:
We know points of inflection will occur when the second derivative changes signs. We can see the second derivative changes signs at x
=−
5
and x
=
3
.
Inflection points:
(
−
5
, f
(
−
5
)
)
→
(−
5,4250
)
(
3
,f
(
3
)
)
→
(
3,1242
)