AP Calculus BC 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative Study Notes - New Syllabus
AP Calculus BC 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative Study Notes- New syllabus
AP Calculus BC 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative 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:
- Connecting a Function, Its First Derivative, and Its Second Derivative
Connecting a Function, Its First Derivative, and Its Second Derivative
Connecting a Function, Its First Derivative, and Its Second Derivative
The behavior of a function \( f(x) \), its first derivative \( f'(x) \), and its second derivative \( f”(x) \) are closely related. Understanding how these three are connected allows us to analyze and sketch functions more effectively.
Important Concepts:
1. Increasing/Decreasing:
- \( f(x) \) is increasing where \( f'(x) > 0 \)
- \( f(x) \) is decreasing where \( f'(x) < 0 \)
2. Local Maximum and Minimum:
- \( f(x) \) has a local maximum where \( f'(x) = 0 \) and \( f”(x) < 0 \)
- \( f(x) \) has a local minimum where \( f'(x) = 0 \) and \( f”(x) > 0 \)
3. Concavity:
- \( f(x) \) is concave up where \( f”(x) > 0 \)
- \( f(x) \) is concave down where \( f”(x) < 0 \)
4. Point of Inflection:
- Occurs where \( f”(x) = 0 \) and changes sign (concavity changes)
Visual Interpretation:
The following table summarizes how the signs of \( f'(x) \) and \( f”(x) \) affect the shape of the graph of \( f(x) \):
| Sign of \( f'(x) \) | Sign of \( f”(x) \) | Behavior of \( f(x) \) |
|---|---|---|
| \( + \) | \( + \) | Increasing and Concave Up |
| \( + \) | \( – \) | Increasing and Concave Down |
| \( – \) | \( + \) | Decreasing and Concave Up |
| \( – \) | \( – \) | Decreasing and Concave Down |
Example:
A function \( f \) has the following properties:
- \( f'(x) > 0 \) for \( x < 1 \) and \( x > 3 \); \( f'(x) < 0 \) for \( 1 < x < 3 \)
- \( f”(x) > 0 \) for \( x < 2 \); \( f”(x) < 0 \) for \( x > 2 \)
Describe the behavior of the function.
▶️Answer/Explanation
- \( f'(x) > 0 \) before \( x = 1 \): function is increasing
- \( f'(x) < 0 \) from \( x = 1 \) to \( x = 3 \): function is decreasing
- \( f'(x) > 0 \) after \( x = 3 \): function is increasing again
- So \( f \) has a local maximum at \( x = 1 \) and local minimum at \( x = 3 \)
- \( f”(x) \) changes sign at \( x = 2 \): inflection point at \( x = 2 \)
- Before \( x = 2 \), function is concave up; after \( x = 2 \), function is concave down
Example:
Given the function \( g(x) = -x^4 + 2x^2 – 1 \), find the interval(s) where \( g \) is concave up and increasing at the same time.
▶️Answer/Explanation
To determine where \( g \) is concave up and increasing:
1. First derivative (to find increasing intervals): \( g'(x) = -4x^3 + 4x = -4x(x^2 – 1) = -4x(x – 1)(x + 1) \)
2. Critical points from \( g'(x) = 0 \): \( x = -1, 0, 1 \)
3. Second derivative (to find concavity): \( g”(x) = -12x^2 + 4 \)
4. Set \( g”(x) > 0 \) for concave up: \( -12x^2 + 4 > 0 \Rightarrow x^2 < \frac{1}{3} \Rightarrow -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \)
5. Now find overlap with intervals where \( g'(x) > 0 \):
From sign analysis, \( g'(x) > 0 \) on: – \( (-1, 0) \) – \( (1, \infty) \)
But concave up only occurs in \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \approx (-0.577, 0.577) \)
So the correct overlap is: \( \left( 0, \frac{1}{\sqrt{3}} \right) \)
Example:
Given the function \( h(x) = x^3 – 2x^2 + x \), find the interval(s) where \( h \) is concave up and decreasing at the same time.
▶️Answer/Explanation
1: First derivative: \( h'(x) = 3x^2 – 4x + 1 \)
2: Second derivative: \( h”(x) = 6x – 4 \)
3: Find intervals of concave up: \( h”(x) > 0 \Rightarrow 6x – 4 > 0 \Rightarrow x > \frac{2}{3} \)
4: Find where \( h'(x) < 0 \) (decreasing): \( h'(x) = 3x^2 – 4x + 1 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16 – 12}}{6} = \frac{4 \pm 2}{6} \Rightarrow x = 1, \frac{1}{3} \)
So, decreasing on \( \left(\frac{1}{3}, 1\right) \)
Now find the overlap: – Concave up: \( x > \frac{2}{3} \) – Decreasing: \( \left( \frac{1}{3}, 1 \right) \)
Overlap is: \( \left( \frac{2}{3}, 1 \right) \)
Example :
In the xy-plane, the graph of the twice-differentiable function \( y = f(x) \) is concave down on the open interval \( (1, 3) \) and is tangent to the line \( y = 4x + 3 \) at \( x = 2 \). Which of the following statements must be true about the derivative of \( f \)?
(A) \( f'(x) \) is constant on the interval \( (2, 2.1) \)
(B) \( f'(x) > 0 \) on the interval \( (2, 2.1) \)
(C) \( f'(x) < 0 \) on the interval \( (2, 2.1) \)
(D) \( f'(x) \geq 4 \) on the interval \( (2, 2.1) \)
(E) \( f'(x) \leq 4 \) on the interval \( (2, 2.1) \)
▶️Answer/Explanation
\( f(x) \) is concave down on \( (1, 3) \)
⇒ \( f”(x) < 0 \) in that interval.
The function is tangent to \( y = 4x + 3 \) at \( x = 2 \) ⇒ \( f'(2) = 4 \)
Since \( f”(x) < 0 \), the slope \( f'(x) \) is decreasing around \( x = 2 \)
Therefore, as \( x \) increases past 2, the slope will decrease: \( f'(x) < f'(2) = 4 \quad \text{on } (2, 2.1) \)
So the correct and guaranteed statement is: (E) \( f'(x) \leq 4 \) on the interval \( (2, 2.1) \)