Grade: Grade 12 Subject: Mathematics Unit: Calculus Introduction Lesson: 6 of 6 SAT: AdvancedMath ACT: Math

Unit Quiz

Test your mastery of calculus fundamentals with this comprehensive unit assessment covering all derivative concepts and applications.

Quiz Instructions

This quiz covers all topics from the Calculus Introduction unit:

  • Introduction to Derivatives (definition, notation, basic rules)
  • Applications of Derivatives (tangent lines, rates of change)
  • Guided Practice (power, product, quotient, and chain rules)
  • Word Problems (motion, optimization, marginal analysis)
  • Common Mistakes (error recognition and prevention)

How to Use This Quiz

  1. Work through all problems without looking at answers
  2. Write down your complete work for each problem
  3. Time yourself: aim to complete in 30-40 minutes
  4. Check your answers only after completing all problems
  5. Review any topics where you made errors

Scoring: 12 questions total. Each question is worth approximately 8 points. A score of 80% or higher indicates mastery.

Quiz Questions

Question 1: Power Rule (8 points)

Find the derivative of f(x) = 4x^5 - 3x^3 + 2x - 7

Show Answer

f'(x) = 20x^4 - 9x^2 + 2

Question 2: Negative and Fractional Exponents (8 points)

Find the derivative of f(x) = 3/x^2 + 2*sqrt(x)

Show Answer

f(x) = 3x^(-2) + 2x^(1/2)

f'(x) = -6x^(-3) + x^(-1/2) = -6/x^3 + 1/sqrt(x)

Question 3: Product Rule (8 points)

Find the derivative of f(x) = (x^2 - 3x)(2x + 5)

Show Answer

f'(x) = (2x - 3)(2x + 5) + (x^2 - 3x)(2)

f'(x) = 4x^2 + 10x - 6x - 15 + 2x^2 - 6x

f'(x) = 6x^2 - 2x - 15

Question 4: Quotient Rule (8 points)

Find the derivative of f(x) = (x^2 + 1)/(x - 2)

Show Answer

f'(x) = [2x(x - 2) - (x^2 + 1)(1)]/(x - 2)^2

f'(x) = [2x^2 - 4x - x^2 - 1]/(x - 2)^2

f'(x) = (x^2 - 4x - 1)/(x - 2)^2

Question 5: Chain Rule (8 points)

Find the derivative of f(x) = (5x^2 - 2x + 1)^3

Show Answer

f'(x) = 3(5x^2 - 2x + 1)^2 * (10x - 2)

f'(x) = 3(10x - 2)(5x^2 - 2x + 1)^2

or: f'(x) = 6(5x - 1)(5x^2 - 2x + 1)^2

Question 6: Combined Rules (8 points)

Find the derivative of f(x) = x^3 * sqrt(x + 4)

Show Answer

Using product rule with chain rule:

f'(x) = 3x^2 * sqrt(x + 4) + x^3 * 1/(2*sqrt(x + 4))

f'(x) = 3x^2 * sqrt(x + 4) + x^3/(2*sqrt(x + 4))

Common denominator: f'(x) = [6x^2(x + 4) + x^3]/(2*sqrt(x + 4))

f'(x) = (7x^3 + 24x^2)/(2*sqrt(x + 4))

Question 7: Tangent Line (8 points)

Find the equation of the tangent line to f(x) = x^3 - 2x at x = 2

Show Answer

f(2) = 8 - 4 = 4, so point is (2, 4)

f'(x) = 3x^2 - 2

f'(2) = 12 - 2 = 10 (slope)

Tangent line: y - 4 = 10(x - 2)

y = 10x - 16

Question 8: Motion Problem (8 points)

A particle moves along a line with position s(t) = t^3 - 12t^2 + 36t meters at time t seconds.

a) Find the velocity function v(t)

b) At what time(s) is the particle at rest?

Show Answer

a) v(t) = s'(t) = 3t^2 - 24t + 36

b) Set v(t) = 0: 3t^2 - 24t + 36 = 0

Divide by 3: t^2 - 8t + 12 = 0

(t - 2)(t - 6) = 0

t = 2 seconds and t = 6 seconds

Question 9: Optimization (8 points)

A farmer wants to fence a rectangular area of 1800 square feet. What dimensions minimize the amount of fencing needed?

Show Answer

Let x = length, y = width. Area: xy = 1800, so y = 1800/x

Perimeter: P = 2x + 2y = 2x + 3600/x

P'(x) = 2 - 3600/x^2 = 0

2 = 3600/x^2

x^2 = 1800

x = sqrt(1800) = 30*sqrt(2) feet (approximately 42.43 feet)

y = 1800/x = 30*sqrt(2) feet

The rectangle is a square with sides of 30*sqrt(2) feet

Question 10: Marginal Analysis (8 points)

A company's profit function is P(x) = -0.5x^2 + 80x - 1500 dollars for x units sold.

a) Find the marginal profit function

b) How many units maximize profit?

Show Answer

a) Marginal profit P'(x) = -x + 80

b) Set P'(x) = 0: -x + 80 = 0

x = 80 units maximize profit

Question 11: Error Detection (8 points)

A student found the derivative of f(x) = sqrt(4x^2 + 9) as follows:

"f'(x) = 1/(2*sqrt(4x^2 + 9))"

Identify the error and give the correct answer.

Show Answer

Error: Forgot to multiply by the derivative of the inner function (chain rule)

Correct work:

f'(x) = 1/(2*sqrt(4x^2 + 9)) * 8x

f'(x) = 4x/sqrt(4x^2 + 9)

Question 12: Conceptual Understanding (8 points)

If f(x) is a function where f'(3) = 0 and f''(3) < 0, what can you conclude about the graph of f at x = 3?

Show Answer

At x = 3:

  • f'(3) = 0 means the tangent line is horizontal (the function has a critical point)
  • f''(3) < 0 means the function is concave down at this point
  • Together, these indicate that x = 3 is a local maximum

The graph of f has a local maximum (peak) at x = 3.

Self-Assessment

After completing the quiz, evaluate your performance:

Score Range Assessment Recommended Action
11-12 correct (92-100%) Excellent - Mastery achieved Ready to proceed to advanced calculus topics
9-10 correct (75-83%) Good - Near mastery Review missed topics, then proceed
7-8 correct (58-67%) Developing - More practice needed Review Guided Practice and Word Problems lessons
0-6 correct (0-50%) Beginning - Significant review needed Restart from Introduction to Derivatives

Next Steps

  • Review any questions you answered incorrectly
  • Revisit the specific lessons for topics where you struggled
  • If you scored 80% or higher, you are ready for the next math unit
  • Consider retaking this quiz after additional practice
  • Explore related topics in the SAT/ACT Skills subject for test preparation