AP Calculus BC 5.6 Determining Concavity of Functions over Their Domains Study Notes - New Syllabus
AP Calculus BC 5.6 Determining Concavity of Functions over Their Domains Study Notes- New syllabus
AP Calculus BC 5.6 Determining Concavity of Functions over Their Domains 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:
- Determining Concavity of Functions over Their Domains
Determining Concavity of Functions over Their Domains
Determining Concavity of Functions over Their Domains
Concavity describes how a function bends: whether it curves upward or downward. To determine where a function is concave up or concave down, we examine the sign of its second derivative \( f”(x) \).
Concavity Rules:
- If \( f”(x) > 0 \) on an interval, then \( f(x) \) is concave up on that interval.
- If \( f”(x) < 0 \) on an interval, then \( f(x) \) is concave down on that interval.
- Points where concavity changes are called inflection points.
Procedure to Determine Concavity:
- Find the second derivative \( f”(x) \).
- Solve \( f”(x) = 0 \) or \( f”(x) \) undefined to find possible inflection points.
- Use test intervals between those points to determine the sign of \( f”(x) \).
Example :
Given the graph of \( f”(x) \) below, determine intervals of concavity and the inflection points.
▶️Answer/Explanation
The graph of \( f”(x) \) touches the x-axis at \( x = -3 \) and \( x = 2 \), and changes sign at these points.
Inflection Points: \( x = -3 \), \( x = 2 \)
Concave up: \( (-\infty, -3) \cup (2, \infty) \)
Concave down: \( (-3, 2) \)
Example :
Analyze the concavity and inflection points for second derivative graph
▶️Answer/Explanation
The graph of \( f”(x) \) crosses the x-axis at \( x = -1 \), \( x = 0 \), and \( x = 3 \).
Inflection Points: \( x = -1 \), \( x = 0 \), \( x = 3 \)
Concave up: \( (-1, 0) \cup (3, \infty) \)
Concave down: \( (-\infty, -1) \cup (0, 3) \)
Example:
Determine the intervals of concavity and inflection points for \( f(x) = x^3 – 6x^2 + 9x \).
▶️Answer/Explanation
First derivative: \( f'(x) = 3x^2 – 12x + 9 \)
Second derivative: \( f”(x) = 6x – 12 \)
\( f”(x) = 0 \): \( 6x – 12 = 0 \Rightarrow x = 2 \)
Test intervals:
On \( (-\infty, 2) \), choose \( x = 1 \): \( f”(1) = 6(1) – 12 = -6 \) → concave down
On \( (2, \infty) \), choose \( x = 3 \): \( f”(3) = 6(3) – 12 = 6 \) → concave up
Conclusion:
Concave down on \( (-\infty, 2) \), concave up on \( (2, \infty) \).
Inflection point at \( x = 2 \). To find the y-value: \( f(2) = 8 – 24 + 18 = 2 \) Inflection point: \( (2, 2) \)
Example:
Find the intervals where \( f(x) = \ln(x^2 + 1) \) is concave up or concave down.
▶️Answer/Explanation
First derivative: \( f'(x) = \frac{2x}{x^2 + 1} \)
Second derivative (using quotient rule): \[ f”(x) = \frac{(x^2 + 1)(2) – 2x(2x)}{(x^2 + 1)^2} = \frac{2(x^2 + 1 – 2x^2)}{(x^2 + 1)^2} = \frac{2(1 – x^2)}{(x^2 + 1)^2} \]
numerator \( = 0 \): \( 1 – x^2 = 0 \Rightarrow x = \pm1 \)
4. Test intervals:
\( (-\infty, -1) \): Choose \( x = -2 \) → \( f”(-2) = \frac{2(1 – 4)}{(4 + 1)^2} = \frac{-6}{25} \) → concave down
\( (-1, 1) \): Choose \( x = 0 \) → \( f”(0) = \frac{2(1)}{1} = 2 \) → concave up
\( (1, \infty) \): Choose \( x = 2 \) → \( f”(2) = \frac{-6}{25} \) → concave down
Example:
Find concavity of the function \( f(x) = e^{-x^2} \).
▶️Answer/Explanation
First derivative (chain rule): \( f'(x) = -2x e^{-x^2} \)
Second derivative: Use product rule: \( f”(x) = (-2)e^{-x^2} + (-2x)(-2x)e^{-x^2} = (-2 + 4x^2)e^{-x^2} \)
Set numerator = 0: \[ -2 + 4x^2 = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}} \]
Test intervals:
\( x = 0 \): \( f”(0) = -2e^{0} = -2 \) → concave down
\( x = 1 \): \( f”(1) = (4 – 2)e^{-1} = 2e^{-1} > 0 \) → concave up
Conclusion:
Concave up on \( \left(-\infty, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, \infty \right) \)
Concave down on \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \)
Inflection points at \( x = \pm \frac{1}{\sqrt{2}} \)