AP Calculus BC 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Study Notes - New Syllabus
AP Calculus BC 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Study Notes- New syllabus
AP Calculus BC 5.4 Using the First Derivative Test to Determine Relative (Local) 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 First Derivative Test to Determine Relative (Local) Extrema
Using the First Derivative Test to Determine Relative (Local) Extrema
Using the First Derivative Test to Determine Relative (Local) Extrema
The First Derivative Test helps identify relative (local) maxima and minima by analyzing how the derivative \( f'(x) \) changes sign around critical points.
Steps:
- Find the derivative \( f'(x) \).
- Determine critical points by solving \( f'(x) = 0 \) or where \( f'(x) \) is undefined.
- Use a sign chart or test values on intervals around the critical points to check the sign of \( f'(x) \).
- Apply the First Derivative Test:
- If \( f'(x) \) changes from positive to negative: local maximum.
- If \( f'(x) \) changes from negative to positive: local minimum.
- If no sign change: no local extremum.
Example:
Use the graph of \( f'(x) \) (the derivative of \( f \)) to determine all local maxima and minima of \( f \).
▶️Answer/Explanation
\( f \) has a local maximum at \( x = -2 \) because \( f'(x) \) changes sign from positive to negative at \( x = -2 \).
\( f \) has a local minimum at \( x = 2 \) because \( f'(x) \) changes sign from negative to positive at \( x = 2 \).
Example :
Use the graph of \( f'(x) \) to determine all relative extrema of \( f \).
▶️Answer/Explanation
\( f \) has a local maximum at \( x = -2.5 \) because \( f'(x) \) changes sign from positive to negative at that point.
\( f \) has a local minimum at \( x = 1 \) because \( f'(x) \) changes sign from negative to positive at \( x = 1 \).
Example:
Use the First Derivative Test to find the local extrema of the function \( f(x) = x^3 – 6x^2 + 9x + 1 \).
▶️Answer/Explanation
Step 1: Find the first derivative:
\( f'(x) = 3x^2 – 12x + 9 \) Step 2: Set \( f'(x) = 0 \):
\( 3x^2 – 12x + 9 = 0 \Rightarrow x^2 – 4x + 3 = 0 \)
\( (x – 1)(x – 3) = 0 \Rightarrow x = 1, 3 \) Step 3: Test intervals around the critical points: – On \( (-\infty, 1) \), pick \( x = 0 \): \( f'(0) = 9 > 0 \) – On \( (1, 3) \), pick \( x = 2 \): \( f'(2) = -3 < 0 \) – On \( (3, \infty) \), pick \( x = 4 \): \( f'(4) = 9 > 0 \) Step 4: Conclusion: – \( f'(x) \) changes from positive to negative at \( x = 1 \): local maximum. – \( f'(x) \) changes from negative to positive at \( x = 3 \): local minimum. Final answer: local max at \( x = 1 \), local min at \( x = 3 \).
Example :
Determine the local extrema of the function \( f(x) = \dfrac{x}{x^2 + 1} \) using the First Derivative Test.
▶️Answer/Explanation
Step 1: Use quotient rule:
\( f'(x) = \dfrac{(x^2 + 1)(1) – x(2x)}{(x^2 + 1)^2} = \dfrac{x^2 + 1 – 2x^2}{(x^2 + 1)^2} = \dfrac{-x^2 + 1}{(x^2 + 1)^2} \) Step 2: Set numerator = 0:
\( -x^2 + 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \) Step 3: Sign chart around \( x = -1 \) and \( x = 1 \): – On \( (-\infty, -1) \), pick \( x = -2 \): numerator = \( -4 + 1 = -3 \) → negative – On \( (-1, 1) \), pick \( x = 0 \): numerator = \( 1 > 0 \) – On \( (1, \infty) \), pick \( x = 2 \): numerator = \( -4 + 1 = -3 \) → negative Step 4: Conclusion: – \( f'(x) \) changes from negative to positive at \( x = -1 \): local min. – \( f'(x) \) changes from positive to negative at \( x = 1 \): local max. Final answer: local min at \( x = -1 \), local max at \( x = 1 \).