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AP Calculus BC
Unit 5: Analytical Applications of Differentiation
5.6 Determining Concavity and Points of Inflection
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Cards (40)
What determines the concavity of a curve?
Second derivative
The graph of a function is concave up when its second derivative is
positive
What is the condition for a graph to be concave down?
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
If f''(x) >0</latex>, the graph is
concave
up.
If
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
, the graph is concave down.
An inflection point occurs where the concavity of a function
changes
What does the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
represent?
Rate of change of slope
Steps to find the second derivative of a function
1️⃣ Find the first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
2️⃣ Differentiate
f
′
(
x
)
f'(x)
f
′
(
x
)
to get
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
What is the first derivative of
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
+
x^{3} +
x
3
+
4
x
2
−
2
x
+
4x^{2} - 2x +
4
x
2
−
2
x
+
5
5
5
?
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
3
x
2
+
3x^{2} +
3
x
2
+
8
x
−
2
8x - 2
8
x
−
2
The second derivative of f(x) = x^{3} + 4x^{2} - 2x + 5</latex> is
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
6
x
+
6x +
6
x
+
8
8
8
A function is concave up when
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
.
A function is concave down when
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
.
A number line is used to visualize the intervals of
concavity
What is the second derivative of a function denoted as?
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
The second derivative represents the rate of change of the slope of the original
function
Steps to find the second derivative
1️⃣ Find the first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
using differentiation rules
2️⃣ Differentiate
f
′
(
x
)
f'(x)
f
′
(
x
)
again to get
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
The first derivative of
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
+
x^{3} +
x
3
+
4
x
2
−
2
x
+
4x^{2} - 2x +
4
x
2
−
2
x
+
5
5
5
is
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
3
x
2
+
3x^{2} +
3
x
2
+
8
x
−
2
8x - 2
8
x
−
2
.
What is the second derivative of f(x) = x^{3} + 4x^{2} - 2x + 5</latex>?
6
x
+
6x +
6
x
+
8
8
8
To determine intervals of concavity, you need to find the second
derivative
If
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
, the function is concave up.
If
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
, what type of concavity does the function have?
Concave down
The function
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
−
6
x
2
+
x^{3} - 6x^{2} +
x
3
−
6
x
2
+
8
x
+
8x +
8
x
+
3
3
3
is concave up for x >2</latex>.
A function's concavity describes the direction its curve
bends
Match the concavity with its corresponding second derivative condition and shape:
Concave Up ↔️
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
,
∪
\cup
∪
Concave Down ↔️
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
,
∩
\cap
∩
The function
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
x^{3}
x
3
is concave up for
x
>
0
x > 0
x
>
0
.
What is an inflection point?
A point where concavity changes
What does a concave up shape look like?
∪
\cup
∪
For
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
x^{3}
x
3
, the second derivative f''(x)</latex> is
6
x
6x
6
x
.
The concavity of a function is determined by its
second
derivative.
For
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
x^{3}
x
3
, where is the function concave down?
x
<
0
x < 0
x
<
0
Steps to find the second derivative
1️⃣ Calculate the first derivative
2️⃣ Differentiate the first derivative again
What is the second derivative of
f
(
x
)
=
f(x) =
f
(
x
)
=
2
x
3
+
2x^{3} +
2
x
3
+
3
x
2
−
5
x
+
3x^{2} - 5x +
3
x
2
−
5
x
+
7
7
7
?
12
x
+
12x +
12
x
+
6
6
6
To find intervals of concavity, you must first find the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
.
The function is concave up when
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
and concave down when
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
.intervals
Match the interval with its concavity for
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
−
6
x
2
+
x^{3} - 6x^{2} +
x
3
−
6
x
2
+
8
x
+
8x +
8
x
+
3
3
3
:
x
<
2
x < 2
x
<
2
↔️ Concave Down
x
>
2
x > 2
x
>
2
↔️ Concave Up
What is a point of inflection?
Change in concavity
For
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
x^{3}
x
3
, the point
(
0
,
0
)
(0,0)
(
0
,
0
)
is a point of inflection.
To find points of inflection, first find the second derivative and solve
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
.concavity
What is the point of inflection for f(x) = x^{3} -6x^{2} + 8x + 3</latex>?
(
2
,
−
1
)
(2, - 1)
(
2
,
−
1
)
Steps to find points of inflection
1️⃣ Find the second derivative
2️⃣ Solve
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
3️⃣ Check concavity before and after each potential point