Unit 5 Practice Test

Analytical Applications of Differentiation - 15 Multiple Choice + 6 Free Response Questions

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Recommended time: 45 minutes

📋 Test Instructions

💡 Important: This practice test covers all topics from Unit 5: Analytical Applications of Differentiation. Take your time and show all your work for free-response questions.
  • Multiple Choice: Select the best answer for each question
  • Free Response: Show all steps and justify your reasoning
  • Time Limit: 45 minutes recommended (AP exam timing)
  • Calculator: Graphing calculator allowed for some questions

📝 Section A: Multiple Choice (15 questions)

1
Find limx→0 (sin(x) - x)/x³ using L'Hôpital's rule.
2
Find limx→∞ (ln(x))/x² using L'Hôpital's rule.
3
For f(x) = x⁴ - 4x³ + 6x² - 4x + 1, find all critical points.
4
Find limx→0 (e^x - 1 - x)/x² using L'Hôpital's rule.
5
For f(x) = x³ - 3x² + 3x - 1, which statement is true?
6
Find limx→0 (1 - cos(x))/x² using L'Hôpital's rule.
7
For f(x) = x⁴ - 8x² + 16, find all inflection points.
8
Find limx→∞ x·e^(-x) using L'Hôpital's rule.
9
For f(x) = x³ - 6x² + 9x + 1, find the absolute maximum on [0, 4].
10
Find limx→0 (x·ln(x)) using L'Hôpital's rule.
11
For f(x) = x⁴ - 4x³ + 6x² - 4x + 1, which statement about concavity is true?
12
Find limx→0 (tan(x) - x)/x³ using L'Hôpital's rule.
13
For f(x) = x³ - 3x + 1, find the value of c guaranteed by Rolle's theorem on [0, 2].
14
Find limx→∞ (x² + 1)^(1/x) using L'Hôpital's rule.
15
For f(x) = x⁴ - 8x² + 16, find the absolute minimum on [-3, 3].

✍️ Section B: Free Response (6 questions)

1
Find limx→0 (sin(x) - x)/x³ using L'Hôpital's rule. Show all steps.

Solution:

1) Direct substitution gives 0/0 (indeterminate form)

2) Apply L'Hôpital's rule: limx→0 (cos(x) - 1)/(3x²)

3) Still 0/0, apply L'Hôpital's rule again: limx→0 (-sin(x))/(6x)

4) Still 0/0, apply L'Hôpital's rule again: limx→0 (-cos(x))/6

5) Evaluate: -cos(0)/6 = -1/6

Answer: -1/6

2
For f(x) = x⁴ - 4x³ + 6x² - 4x + 1, find all critical points and classify each as a local maximum, local minimum, or neither.

Solution:

1) Find f'(x) = 4x³ - 12x² + 12x - 4 = 4(x³ - 3x² + 3x - 1)

2) Factor: f'(x) = 4(x - 1)³

3) Set f'(x) = 0: 4(x - 1)³ = 0, so x = 1

4) Find f''(x) = 12x² - 24x + 12 = 12(x² - 2x + 1) = 12(x - 1)²

5) At x = 1: f''(1) = 12(0)² = 0

6) Since f''(1) = 0, the second derivative test is inconclusive

7) Check first derivative test: f'(x) = 4(x - 1)³ changes sign at x = 1

8) For x < 1: f'(x) < 0 (decreasing)

9) For x > 1: f'(x) > 0 (increasing)

10) Therefore, x = 1 is a local minimum

Answer: x = 1 is a local minimum

3
Find limx→∞ (ln(x))/x² using L'Hôpital's rule.

Solution:

1) Direct substitution gives ∞/∞ (indeterminate form)

2) Apply L'Hôpital's rule: limx→∞ (1/x)/(2x)

3) Simplify: limx→∞ 1/(2x²)

4) Evaluate: 1/∞ = 0

Answer: 0

4
For f(x) = x³ - 3x² + 3x - 1, find all inflection points and intervals of concavity.

Solution:

1) Find f'(x) = 3x² - 6x + 3 = 3(x² - 2x + 1) = 3(x - 1)²

2) Find f''(x) = 6x - 6 = 6(x - 1)

3) Set f''(x) = 0: 6(x - 1) = 0, so x = 1

4) Test intervals:

- For x < 1: f''(x) < 0, so f is concave down on (-∞, 1)

- For x > 1: f''(x) > 0, so f is concave up on (1, ∞)

5) Since f'' changes sign at x = 1, there is an inflection point at x = 1

6) Find y-coordinate: f(1) = 1 - 3 + 3 - 1 = 0

Answer: Inflection point at (1, 0); concave down on (-∞, 1), concave up on (1, ∞)

5
Find limx→0 (1 - cos(x))/x² using L'Hôpital's rule.

Solution:

1) Direct substitution gives 0/0 (indeterminate form)

2) Apply L'Hôpital's rule: limx→0 sin(x)/(2x)

3) Still 0/0, apply L'Hôpital's rule again: limx→0 cos(x)/2

4) Evaluate: cos(0)/2 = 1/2

Answer: 1/2

6
For f(x) = x⁴ - 8x² + 16, find the absolute maximum and minimum on [-3, 3].

Solution:

1) Find f'(x) = 4x³ - 16x = 4x(x² - 4) = 4x(x - 2)(x + 2)

2) Critical points: x = 0, x = ±2

3) Check endpoints and critical points:

- f(-3) = 81 - 72 + 16 = 25

- f(-2) = 16 - 32 + 16 = 0

- f(0) = 0 - 0 + 16 = 16

- f(2) = 16 - 32 + 16 = 0

- f(3) = 81 - 72 + 16 = 25

4) Absolute maximum: 25 at x = ±3

5) Absolute minimum: 0 at x = ±2

Answer: Absolute maximum = 25 at x = ±3; Absolute minimum = 0 at x = ±2

📊 Test Results

📚 Study Recommendations

If you scored 80%+:
  • Great job! You have a solid foundation
  • Focus on the few topics you missed
  • Move on to Unit 6: Integration and Accumulation
  • Practice more challenging L'Hôpital's rule problems
If you scored below 80%:
  • Review the concepts you struggled with
  • Practice more L'Hôpital's rule applications
  • Focus on function analysis techniques
  • Use the main Unit 5 page for review