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AP Calculus AB
Unit 5: Analytical Applications of Differentiation
5.9 Connecting <latex>f'(x)</latex> and <latex>f''(x)</latex> with the Graph of <latex>f(x)</latex>
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Cards (121)
When
f
(
x
)
f(x)
f
(
x
)
is increasing,
f
′
(
x
)
f'(x)
f
′
(
x
)
is positive
Match the behavior of
f
(
x
)
f(x)
f
(
x
)
with the sign of
f
′
(
x
)
f'(x)
f
′
(
x
)
:
Increasing ↔️ Positive
Decreasing ↔️ Negative
Constant ↔️ Zero
What does the derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
represent in calculus?
Rate of change of
f
(
x
)
f(x)
f
(
x
)
What is the value of
f
′
(
x
)
f'(x)
f
′
(
x
)
when
f
(
x
)
f(x)
f
(
x
)
is constant?
Zero
If
f
′
(
x
)
>
0
f'(x) > 0
f
′
(
x
)
>
0
, the graph of
f
(
x
)
f(x)
f
(
x
)
has an upward slope.
True
What is a point of inflection on the graph of
f
(
x
)
f(x)
f
(
x
)
?
Where
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
changes sign
If
f
′
(
x
)
f'(x)
f
′
(
x
)
is positive,
f
(
x
)
f(x)
f
(
x
)
is increasing.
True
When
f
′
(
x
)
f'(x)
f
′
(
x
)
is positive, f(x)</latex> is increasing.
True
Match the behavior of
f
(
x
)
f(x)
f
(
x
)
with the sign of
f
′
(
x
)
f'(x)
f
′
(
x
)
:
Increasing ↔️ Positive
Decreasing ↔️ Negative
Constant ↔️ Zero
If
f
(
x
)
f(x)
f
(
x
)
is increasing,
f
′
(
x
)
f'(x)
f
′
(
x
)
is positive.
True
If
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
, the graph of
f
(
x
)
f(x)
f
(
x
)
is concave down.
True
When
f
′
(
x
)
>
0
f'(x) > 0
f
′
(
x
)
>
0
,
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
, the graph of
f
(
x
)
f(x)
f
(
x
)
is upward sloping and concave
If
f
′
(
x
)
>
0
f'(x) > 0
f
′
(
x
)
>
0
in an interval (a, b)</latex>, what is
f
(
x
)
f(x)
f
(
x
)
doing in that interval?
Increasing
What is a critical point of
f
(
x
)
f(x)
f
(
x
)
?
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
or undefined
What does
f
′
(
x
)
f'(x)
f
′
(
x
)
represent at a given point?
Rate of change of
f
(
x
)
f(x)
f
(
x
)
If
f
(
x
)
f(x)
f
(
x
)
is an increasing function, then
f
′
(
x
)
f'(x)
f
′
(
x
)
is positive
What is a point of inflection on the graph of
f
(
x
)
f(x)
f
(
x
)
?
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
and changes sign
If
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
is positive, the graph of
f
(
x
)
f(x)
f
(
x
)
is concave up.
True
If
f
′
(
x
)
f'(x)
f
′
(
x
)
is negative, the graph of
f
(
x
)
f(x)
f
(
x
)
is decreasing.
True
If
f
′
(
x
)
f'(x)
f
′
(
x
)
changes from positive to negative at a critical point, it indicates a local maximum.
True
If f''(x) < 0</latex>, the graph of
f
(
x
)
f(x)
f
(
x
)
is concave down.
True
If
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
is positive, the concavity of f(x)</latex> is concave up
For
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
−
6
x
2
+
x^{3} - 6x^{2} +
x
3
−
6
x
2
+
5
5
5
, what is
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
?
6
x
−
12
6x - 12
6
x
−
12
At
x
=
x =
x
=
2
2
2
,
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
, indicating a point of inflection for f(x)</latex>.
True
Inflection points occur where
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
is zero or undefined
Steps to find inflection points for
f
(
x
)
f(x)
f
(
x
)
1️⃣ Find the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
2️⃣ Set
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
3️⃣ Solve for
x
x
x
4️⃣ Analyze concavity around
x
x
x
5️⃣ Identify inflection points
What sign change in
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
indicates an inflection point?
Positive to negative
If
f
′
(
x
)
<
0
f'(x) < 0
f
′
(
x
)
<
0
, the graph of
f
(
x
)
f(x)
f
(
x
)
is downward sloping.
True
The derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
represents the rate of change of the function
f
(
x
)
f(x)
f
(
x
)
at a given point
Summarize the relationship between f(x)</latex>,
f
′
(
x
)
f'(x)
f
′
(
x
)
, and the graph of
f
(
x
)
f(x)
f
(
x
)
:
1️⃣ If
f
′
(
x
)
f'(x)
f
′
(
x
)
is positive,
f
(
x
)
f(x)
f
(
x
)
is increasing and the graph slopes upward.
2️⃣ If
f
′
(
x
)
f'(x)
f
′
(
x
)
is negative,
f
(
x
)
f(x)
f
(
x
)
is decreasing and the graph slopes downward.
3️⃣ If
f
′
(
x
)
f'(x)
f
′
(
x
)
is zero,
f
(
x
)
f(x)
f
(
x
)
is constant and the graph is a horizontal line.
Match the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
with the concavity of
f
(
x
)
f(x)
f
(
x
)
:
Positive ↔️ Concave up
Negative ↔️ Concave down
Zero ↔️ Possible point of inflection
Arrange the steps to determine concavity and points of inflection using
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
:
1️⃣ Find the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
.
2️⃣ Set
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
to find possible points of inflection.
3️⃣ Determine the sign of
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
in intervals around critical points.
4️⃣ Identify concavity as concave up if
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
and concave down if
f
′
′
(
x
)
<
0
f''(x) < 0
f
′′
(
x
)
<
0
.
5️⃣ If
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
changes sign, it indicates a point of inflection.
Match the derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
with the behavior of
f
(
x
)
f(x)
f
(
x
)
:
Positive ↔️ Increasing
Negative ↔️ Decreasing
Zero ↔️ Constant
If
f
′
(
x
)
f'(x)
f
′
(
x
)
is zero at
x
=
x =
x
=
c
c
c
, then
f
(
x
)
f(x)
f
(
x
)
is momentarily constant
Match the behavior of
f
′
(
x
)
f'(x)
f
′
(
x
)
at
x
=
x =
x
=
c
c
c
with the type of extremum:
Positive to negative ↔️ Local maximum
Negative to positive ↔️ Local minimum
No change in sign ↔️ Not an extremum
If
f
′
(
x
)
f'(x)
f
′
(
x
)
changes from negative to positive at x = c</latex>, then
f
(
x
)
f(x)
f
(
x
)
has a local minimum at
x
=
x =
x
=
c
c
c
.
True
Match the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
with the concavity of the graph of
f
(
x
)
f(x)
f
(
x
)
:
Positive ↔️ Concave up
Negative ↔️ Concave down
Zero ↔️ Possible point of inflection
If
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
, the graph of
f
(
x
)
f(x)
f
(
x
)
is concave up
A point of inflection occurs when
f
′
′
(
x
)
=
f''(x) =
f
′′
(
x
)
=
0
0
0
and the concavity of
f
(
x
)
f(x)
f
(
x
)
changes
What happens to the concavity of
f
(
x
)
f(x)
f
(
x
)
at
x
=
x =
x
=
2
2
2
for
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
−
6
x
2
+
x^{3} - 6x^{2} +
x
3
−
6
x
2
+
5
5
5
?
Changes concavity
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