AP Calculus BC 5.7 Using the Second Derivative Test to Determine Extrema Study Notes - New Syllabus
AP Calculus BC 5.7 Using the Second Derivative Test to Determine Extrema Study Notes- New syllabus
AP Calculus BC 5.7 Using the Second Derivative Test to Determine Extrema Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- A function’s derivative can be used to understand some behaviors of the function.
Key Concepts:
- Using the Second Derivative Test to Determine Extrema
Using the Second Derivative Test to Determine Extrema
Using the Second Derivative Test to Determine Extrema
The Second Derivative Test is used to determine whether a critical point of a function is a local maximum, local minimum, or neither.
Steps:
- Find the critical points by solving \( f'(x) = 0 \).
- Evaluate the second derivative \( f”(x) \) at those critical points.
Interpretation:
- If \( f”(c) > 0 \), the function has a local minimum at \( x = c \).
- If \( f”(c) < 0 \), the function has a local maximum at \( x = c \).
- If \( f”(c) = 0 \), the test is inconclusive. Use another method like the First Derivative Test.
Example:
Let \( f(x) = x^3 – 3x^2 + 4 \). Use the second derivative test to classify the critical points.
▶️Answer/Explanation
\( f'(x) = 3x^2 – 6x \)
Set \( f'(x) = 0 \):
\( 3x^2 – 6x = 0 \Rightarrow 3x(x – 2) = 0 \Rightarrow x = 0, 2 \)
\( f”(x) = 6x – 6 \)
Evaluate at each critical point:
- \( f”(0) = 6(0) – 6 = -6 \Rightarrow \) Local maximum at \( x = 0 \)
- \( f”(2) = 6(2) – 6 = 6 \Rightarrow \) Local minimum at \( x = 2 \)
Example:
Consider \( f(x) = x^4 – 4x^2 \). Classify the critical points using the second derivative test.
▶️Answer/Explanation
\( f'(x) = 4x^3 – 8x \)
Set \( f'(x) = 0 \):
\( 4x(x^2 – 2) = 0 \Rightarrow x = 0, \pm \sqrt{2} \)
Second derivative: \( f”(x) = 12x^2 – 8 \)
- \( f”(0) = -8 \Rightarrow \) Local maximum at \( x = 0 \)
- \( f”(\sqrt{2}) = 12(2) – 8 = 16 \Rightarrow \) Local minimum at \( x = \sqrt{2} \)
- \( f”(-\sqrt{2}) = 12(2) – 8 = 16 \Rightarrow \) Local minimum at \( x = -\sqrt{2} \)
Example:
The position of a particle moving along a straight line is given by the function \( s(t) = t^3 – 6t^2 + 9t \), where \( s \) is in meters and \( t \) is in seconds.
Use the Second Derivative Test to determine whether the particle’s position is at a local maximum or minimum.
▶️Answer/Explanation
velocity function
\( v(t) = s'(t) = 3t^2 – 12t + 9 \)
Set \( v(t) = 0 \):
\( 3t^2 – 12t + 9 = 0 \Rightarrow t^2 – 4t + 3 = 0 \Rightarrow (t – 1)(t – 3) = 0 \)
Critical points: \( t = 1 \), \( t = 3 \)
acceleration (second derivative)
\( a(t) = v'(t) = s”(t) = 6t – 12 \)
Evaluate at each critical point:
- \( a(1) = 6(1) – 12 = -6 \Rightarrow \) Local maximum at \( t = 1 \)
- \( a(3) = 6(3) – 12 = 6 \Rightarrow \) Local minimum at \( t = 3 \)
Conclusion:
- The particle reaches a local maximum position at \( t = 1 \) seconds.
- The particle reaches a local minimum position at \( t = 3 \) seconds.