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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 (45)
What does a negative second derivative at a critical point indicate?
Relative maximum
Steps to find the critical points of a function f(x)
1️⃣ Compute the first derivative f'(x)
2️⃣ Set f'(x) = 0 and solve for x
3️⃣ Check where f'(x) is undefined
4️⃣ List all x values
For f(x) = x² - 4x + 3, what is f'(x)?
2x - 4
If f''(x) < 0 at a critical point, the function has a relative
maximum
What type of extremum does the critical point x = 2 have for f(x) = x² - 4x + 3?
Relative minimum
For f(x) = x² - 4x + 3, the first derivative f'(x) is
2x
The second derivative is calculated by differentiating the
first
derivative.
True
The second derivative measures the rate of change of the
slope
Given f(x) = x^3 - 3x^2 + 5x - 2, what is f'(x)?
3x^2 - 6x + 5
The Second Derivative Test can determine whether a critical point is a relative maximum, minimum, or inconclusive.
True
If f''(2) = 6 > 0, then x = 2 is a relative
minimum
What is the first derivative of the function f(x) = x^2 - 4x + 3?
f'(x) = 2x - 4
What is the only critical point of the function f(x) = x^2 - 4x + 3?
x = 2
Steps to calculate the second derivative
1️⃣ Find the first derivative f'(x)
2️⃣ Differentiate f'(x) to get f''(x)
What is the value of f''(2) for the function f(x) = x^3 - 3x^2 + 5x - 2?
f''(2) = 6
If f''(x) > 0 at a critical point, it indicates a relative maximum.
False
To find the critical points of a function, we need to find the values of x where the first derivative f'(x) is zero or
undefined
Critical points occur only where f'(x) = 0.
False
A positive second derivative at a critical point indicates a relative minimum.
True
Steps to find the critical points of a function f(x)
1️⃣ Find f'(x)
2️⃣ Set f'(x) = 0 and solve for x
3️⃣ Check where f'(x) is undefined
4️⃣ List all critical points
Match the condition with the relative extremum:
f''(x) < 0 ↔️ Relative maximum
f''(x) > 0 ↔️ Relative minimum
f''(x) = 0 ↔️ Test inconclusive
Critical points are found by setting the first derivative equal to
zero
The second derivative of f(x) = x^3 - 3x^2 + 5x - 2 is f''(x) =
6x - 6
Match the condition with the relative extremum:
f''(x) < 0 ↔️ Relative maximum
f''(x) > 0 ↔️ Relative minimum
f''(x) = 0 ↔️ Inconclusive
Steps to find the critical points of a function
1️⃣ Find the first derivative f'(x)
2️⃣ Set f'(x) = 0 and solve for x
3️⃣ Identify undefined points
4️⃣ List critical points
To find the critical points, the first derivative must be set equal to
zero
The Second Derivative Test is used to determine whether a critical point is a relative
maximum
What does it mean if f''(x) = 0 at a critical point?
Test is inconclusive
What is the purpose of the Second Derivative Test?
Determine relative extrema
For f(x) = x² - 4x + 3, what is f''(x)?
2
What is the critical value for the function f(x) = x² - 4x + 3?
x = 2
What does the second derivative measure?
Concavity of the function
You must check where the first derivative is undefined to find all
critical points
.
True
Steps to calculate the second derivative
1️⃣ Find the first derivative f'(x)
2️⃣ Differentiate f'(x) to get f''(x)
What is the first derivative of f(x) = x^3 - 3x^2 + 5x - 2?
3x^2 - 6x + 5
If f'(x) is defined for all x, then the only critical point of f(x) = x^2 - 4x + 3 is x =
2
The first derivative f'(x) = 2x - 4 is defined for all x.
True
What is the first derivative of f(x) = x^3 - 3x^2 + 5x - 2?
f'(x) = 3x^2 - 6x + 5
Since f''(2) = 6 > 0, the function has a
relative minimum
at x = 2.
True
Steps to determine whether a critical point is a relative maximum or minimum using the Second Derivative Test
1️⃣ Calculate f''(x)
2️⃣ Evaluate f''(x) at each critical point
3️⃣ Interpret the sign of f''(x)
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