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Unit 5: Analytical Applications of Differentiation
5.5 Using the Second Derivative Test for Relative Extrema
Applying the Second Derivative Test:
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Cards (58)
What does the Second Derivative Test determine about a function at a critical point?
Relative extrema
If
f
′
′
(
c
)
>
0
f''(c) > 0
f
′′
(
c
)
>
0
, then
f
(
x
)
f(x)
f
(
x
)
has a local minimum at
x
=
x =
x
=
c
c
c
.
True
The derivative of
x
n
x^{n}
x
n
is nx^{n - 1}.
What is the Second Derivative Test used for?
Determining relative extrema
If
f
′
′
(
c
)
=
f''(c) =
f
′′
(
c
)
=
0
0
0
, the Second Derivative Test is inconclusive
What does the first derivative of a function represent?
Slope of the tangent line
What is the first derivative of
f
(
x
)
=
f(x) =
f
(
x
)
=
2
x
3
−
5
x
2
+
2x^{3} - 5x^{2} +
2
x
3
−
5
x
2
+
7
x
−
3
7x - 3
7
x
−
3
?
6
x
2
−
10
x
+
6x^{2} - 10x +
6
x
2
−
10
x
+
7
7
7
What does the second derivative of a function represent?
Rate of change of the slope
Steps to find critical points of a function
1️⃣ Compute the first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
2️⃣ Set
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
3️⃣ Solve for
x
x
x
To determine the nature of extrema, you evaluate the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
at each critical point
What should you do if
f
′
′
(
c
)
=
f''(c) =
f
′′
(
c
)
=
0
0
0
?
Use the First Derivative Test
What does the first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
of a function represent?
Slope of the tangent line
Steps to apply the Second Derivative Test
1️⃣ Find critical points by solving
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
2️⃣ Compute the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
3️⃣ Evaluate
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
at each critical point
4️⃣ Determine relative extrema based on the sign of
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
What does the first derivative of a function represent?
Slope of the tangent line
If f''(c) > 0</latex>, what type of relative extremum does
f
(
x
)
f(x)
f
(
x
)
have at
x
=
x =
x
=
c
c
c
?
Local minimum
If
f
′
′
(
c
)
>
0
f''(c) > 0
f
′′
(
c
)
>
0
, then
f
(
x
)
f(x)
f
(
x
)
has a local minimum at
x
=
x =
x
=
c
c
c
True
What does the Second Derivative Test use to identify extrema?
Value of
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
Match the function with its derivative:
s
i
n
(
x
)
sin(x)
s
in
(
x
)
↔️
c
o
s
(
x
)
cos(x)
cos
(
x
)
c
o
s
(
x
)
cos(x)
cos
(
x
)
↔️
−
s
i
n
(
x
)
- sin(x)
−
s
in
(
x
)
e
x
e^{x}
e
x
↔️
e
x
e^{x}
e
x
l
n
(
x
)
ln(x)
l
n
(
x
)
↔️
1
/
x
1 / x
1/
x
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 find
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
To find the critical points of a function, set the first derivative equal to
zero
How do you identify the critical points of a function
f
(
x
)
f(x)
f
(
x
)
?
Solve
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
If
f
′
′
(
c
)
<
0
f''(c) < 0
f
′′
(
c
)
<
0
, then f(x)</latex> has a local maximum
The Second Derivative Test becomes inconclusive when f''(c) =
0
What is the derivative of
s
i
n
(
x
)
sin(x)
s
in
(
x
)
?
c
o
s
(
x
)
cos(x)
cos
(
x
)
The first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
is derived from the original function
f
(
x
)
f(x)
f
(
x
)
.
True
Steps to find critical points of a function
f
(
x
)
f(x)
f
(
x
)
1️⃣ Compute the first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
.
2️⃣ Set
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
.
3️⃣ Solve for
x
x
x
to find critical points.
What do you set the first derivative equal to in order to find critical points?
Zero
The Second Derivative Test evaluates the
second
derivative at critical points.
What is the value of
f
′
′
(
1
)
f''(1)
f
′′
(
1
)
for
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
−
6
x
2
+
x^{3} - 6x^{2} +
x
3
−
6
x
2
+
9
x
+
9x +
9
x
+
1
1
1
?
−
6
- 6
−
6
If
f
′
′
(
c
)
<
0
f''(c) < 0
f
′′
(
c
)
<
0
, then
f
(
x
)
f(x)
f
(
x
)
has a local maximum at
x
=
x =
x
=
c
c
c
.
Steps to apply the Second Derivative Test
1️⃣ Evaluate the second derivative
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
at each critical point c</latex>.
2️⃣ Check the sign of
f
′
′
(
c
)
f''(c)
f
′′
(
c
)
to determine the type of extrema.
The second derivative describes the rate of change of the
slope
The second derivative is derived from the first
derivative
Critical points are the values of
x
x
x
that satisfy the equation f'(x) = 0
Steps to find critical points of a function
f
(
x
)
f(x)
f
(
x
)
1️⃣ Compute the first derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
.
2️⃣ Set
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
.
3️⃣ Solve the equation
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
0
0
for
x
x
x
.
Under what condition is the Second Derivative Test inconclusive?
f
′
′
(
c
)
=
f''(c) =
f
′′
(
c
)
=
0
0
0
Match the critical point with the type of extrema based on the Second Derivative Test:
x
=
x =
x
=
1
1
1
↔️ Local maximum
x
=
x =
x
=
3
3
3
↔️ Local minimum
Which test should you use if
f
′
′
(
c
)
=
f''(c) =
f
′′
(
c
)
=
0
0
0
?
First Derivative Test
The Second Derivative Test identifies extrema based on the sign of
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
at critical points.
The Second Derivative Test is used to determine whether a function has a local maximum or
minimum
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