AP Calculus BC 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing Study Notes - New Syllabus
AP Calculus BC 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing Study Notes- New syllabus
AP Calculus BC 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing 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 Intervals on Which a Function Is Increasing or Decreasing
Determining Intervals on Which a Function Is Increasing or Decreasing
Determining Intervals on Which a Function Is Increasing or Decreasing
To determine where a function \( f(x) \) is increasing or decreasing, we analyze the sign of its derivative \( f'(x) \).
Key Steps:
- Find the first derivative \( f'(x) \).
- Identify critical points by solving \( f'(x) = 0 \) or where \( f'(x) \) is undefined.
- Divide the domain into intervals around the critical points.
- Pick test points in each interval and determine the sign of \( f'(x) \).
- If \( f'(x) > 0 \) on an interval, then \( f(x) \) is increasing there.
- If \( f'(x) < 0 \) on an interval, then \( f(x) \) is decreasing there.
Example :
The graph of \( f'(x) \) is shown. Identify the intervals where \( f(x) \) is increasing or decreasing. Justify your answer.
Increasing:
Decreasing:
▶️Answer/Explanation
Based on the graph of \( f'(x) \):
\( f'(x) > 0 \) on \( (-1, 0) \) and \( (3, \infty) \), so \( f(x) \) is increasing there.
\( f'(x) < 0 \) on \( (-\infty, -1) \) and \( (0, 3) \), so \( f(x) \) is decreasing there.
Example :
The graph of \( f'(x) \) is shown. Identify the intervals where \( f(x) \) is increasing or decreasing. Justify your answer.
Increasing:
Decreasing:
▶️Answer/Explanation
We use the rule that a function \( f(x) \) is increasing when \( f'(x) > 0 \) and decreasing when \( f'(x) < 0 \).
From the graph:
\( f'(x) > 0 \) on \( (-\infty, -1.5) \) and \( (-0.5, \infty) \), so \( f(x) \) is increasing there.
\( f'(x) < 0 \) on \( (-1.5, -0.5) \), so \( f(x) \) is decreasing there.
Example :
Determine the intervals on which \( f(x) = x^3 – 6x^2 + 9x \) is increasing or decreasing.
▶️Answer/Explanation
\( f'(x) = 3x^2 – 12x + 9 \)
\( f'(x) = 0 \):
\( 3x^2 – 12x + 9 = 0 \Rightarrow x^2 – 4x + 3 = 0 \Rightarrow (x – 1)(x – 3) = 0 \)
Critical points: \( x = 1, 3 \)
Test intervals: \( (-\infty, 1) \), \( (1, 3) \), \( (3, \infty) \)
- Pick \( x = 0 \): \( f'(0) = 9 > 0 \) → Increasing
- Pick \( x = 2 \): \( f'(2) = 3(4) – 12(2) + 9 = 12 – 24 + 9 = -3 \) → Decreasing
- Pick \( x = 4 \): \( f'(4) = 3(16) – 48 + 9 = 48 – 48 + 9 = 9 > 0 \) → Increasing
Example :
For the function \( f(x) = \ln(x^2 + 1) \), determine the intervals of increase and decrease.
▶️Answer/Explanation
\( f'(x) = \frac{2x}{x^2 + 1} \)
\( f'(x) = 0 \):
\( \frac{2x}{x^2 + 1} = 0 \Rightarrow x = 0 \)
Test intervals: \( (-\infty, 0) \), \( (0, \infty) \)
- Pick \( x = -1 \): \( f'(-1) = \frac{-2}{2} = -1 \) → Decreasing
- Pick \( x = 1 \): \( f'(1) = \frac{2}{2} = 1 \) → Increasing
Example:
A particle \( X \) moves along the positive x-axis so that its position at time \( t \geq 0 \) is given by \( x(t) = 2t^3 – 4t^2 + 4 \).
(a) Is particle \( X \) moving toward the left or toward the right at time \( t = 2 \)? Give a reason for your answer.
(b) At what time \( t \geq 0 \) is particle \( X \) farthest to the left? Justify your answer.
(c) A second particle \( Y \) moves along the positive y-axis so that its position at time \( t \) is given by \( y(t) = 4t + 5 \). The origin and the positions of particles \( X \) and \( Y \) are vertices of a rectangle in the first quadrant.
Find the rate of change of the area of the rectangle at time \( t = 2 \). Show the work that leads to your answer.
▶️Answer/Explanation
(a)
To determine the direction of motion, we examine the velocity \( x'(t) \), which is the derivative of the position function:
\( x'(t) = \frac{d}{dt}(2t^3 – 4t^2 + 4) = 6t^2 – 8t \)
Evaluate at \( t = 2 \):
\( x'(2) = 6(2)^2 – 8(2) = 24 – 16 = 8 \)
Since \( x'(2) > 0 \), the particle is moving toward the right at \( t = 2 \).
(b)
The particle is farthest left when \( x(t) \) is at a minimum. This occurs at a critical point where \( x'(t) = 0 \).
We already have:
\( x'(t) = 6t^2 – 8t = 2t(3t – 4) \)
Set derivative equal to 0:
\( 2t(3t – 4) = 0 \Rightarrow t = 0 \) or \( t = \frac{4}{3} \)
Now check concavity using second derivative:
\( x”(t) = \frac{d}{dt}(6t^2 – 8t) = 12t – 8 \)
- At \( t = 0 \), \( x”(0) = -8 \Rightarrow \) concave down → local maximum
- At \( t = \frac{4}{3} \), \( x”\left(\frac{4}{3}\right) = 12\left(\frac{4}{3}\right) – 8 = 16 – 8 = 8 \Rightarrow \) concave up → local minimum
Therefore, particle \( X \) is farthest to the left at \( t = \frac{4}{3} \).
(c)
The area of the rectangle is given by:
\( A(t) = x(t) \cdot y(t) \)
Use the product rule to differentiate:
\( A'(t) = x'(t)y(t) + x(t)y'(t) \)
At \( t = 2 \):
- \( x(2) = 2(8) – 4(4) + 4 = 16 – 16 + 4 = 4 \)
- \( x'(2) = 6(4) – 8(2) = 24 – 16 = 8 \)
- \( y(2) = 4(2) + 5 = 13 \)
- \( y'(t) = 4 \) (constant)
Now compute:
\( A'(2) = x'(2)y(2) + x(2)y'(2) = 8(13) + 4(4) = 104 + 16 = 120 \)
So, the rate of change of the area at \( t = 2 \) is 120 units² per unit time.