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Quizzes > High School Quizzes > Mathematics

Unit 5 AP Calc AB Practice Quiz

Enhance your AP Calculus with targeted practice

Difficulty: Moderate
Grade: Grade 12
Study OutcomesCheat Sheet
Colorful paper art promoting Unit 5 Calculus Challenge trivia quiz for high school students.

What is the definition of the derivative f'(x) using limit notation?
lim (h→0) (f(x+h) - f(x)) / h
lim (h→0) (f(x) - f(x+h)) / h
lim (x→∞) (f(x+h) - f(x)) / h
lim (h→0) (f(x+h) + f(x)) / h
The derivative is defined as the limit as h approaches 0 of (f(x+h) - f(x)) divided by h. This expression gives the instantaneous rate of change and the slope of the tangent line at x.
Which of the following is an antiderivative of 2x?
x^2 + C
x^2
2x + C
x + C
An antiderivative of 2x is a function whose derivative equals 2x. Since the derivative of x^2 is 2x, the general antiderivative is x^2 + C, where C is an arbitrary constant.
What is the derivative of x^3 using the power rule?
3x^2
x^2
3x
x^2 + C
Applying the power rule, which states that the derivative of x^n is n*x^(n-1), we find that the derivative of x^3 is 3x^2. This fundamental rule is essential for differentiating polynomial functions.
According to the Fundamental Theorem of Calculus, if F(x) = ∫[a,x] f(t) dt, what is F'(x)?
f(x)
∫[a,x] f(t) dt
f(a)
0
The Fundamental Theorem of Calculus part 1 tells us that if F(x) is defined as the integral from a constant a to x of f(t), then the derivative F'(x) is simply f(x). This directly links the process of integration and differentiation.
What is the limit of the constant function f(x) = 3 as x approaches any value?
3
0
Infinity
Depends on x
Since f(x) = 3 is a constant function, its value remains 3 regardless of the input. Therefore, the limit as x approaches any value is simply 3.
Using the chain rule, what is the derivative of f(x) = (2x + 3)^4?
8(2x + 3)^3
4(2x + 3)^3
8(2x + 3)^4
4(2x + 3)^4
The chain rule tells us to differentiate the outer function and then multiply by the derivative of the inner function. Differentiating (2x + 3)^4 produces 4(2x + 3)^3 and multiplying by the derivative of (2x + 3), which is 2, gives 8(2x + 3)^3.
What is the derivative of f(x) = x^2 · sin(x) using the product rule?
2x sin(x) + x^2 cos(x)
2x sin(x) - x^2 cos(x)
x^2 cos(x)
sin(x) + cos(x)
The product rule states that the derivative of two multiplied functions is f'(x)g(x) + f(x)g'(x). Here, differentiating x^2 gives 2x and differentiating sin(x) gives cos(x), yielding 2x sin(x) + x^2 cos(x).
Find the derivative of f(x) = (e^x) / x using the quotient rule.
e^x (x - 1) / x^2
e^x (1 - x) / x^2
e^x / x
e^x (x + 1) / x^2
Applying the quotient rule, where the derivative is given by (g'(x)h(x) - g(x)h'(x)) / [h(x)]^2, and letting g(x) = e^x and h(x) = x, we arrive at e^x (x - 1) / x^2 after simplification.
Evaluate the definite integral ∫ from 1 to 3 of 2 dx.
4
2
6
8
The integral of a constant function 2 over the interval [1,3] equals 2 times the length of the interval, which is 2 (since 3 - 1 = 2). Multiplying 2 by 2 gives 4, representing the area under the constant function.
Find the indefinite integral ∫ 3x^2 dx.
x^3 + C
3x + C
x^2 + C
3x^3 + C
Using the power rule for integration, ∫ x^n dx = x^(n+1)/(n+1) + C. For 3x^2, this yields 3·(x^3/3) + C which simplifies to x^3 + C.
If F(x) is an antiderivative of f(x), what does the Fundamental Theorem of Calculus state for ∫ from a to b of f(x) dx?
F(b) - F(a)
F(a) - F(b)
F(b) + F(a)
F(b) * F(a)
The Fundamental Theorem of Calculus connects antiderivatives and definite integrals by stating that the integral from a to b of f(x) dx equals F(b) - F(a), where F is any antiderivative of f. This theorem is a cornerstone of calculus.
What is the formula for the average value of a continuous function f(x) on the interval [a, b]?
(1/(b-a)) ∫ from a to b of f(x) dx
(b-a) ∫ from a to b of f(x) dx
∫ from a to b of f(x) dx
(1/2) ∫ from a to b of f(x) dx
The average value of a function on an interval is found by dividing the total area under its curve by the length of the interval. Mathematically, this is expressed as (1/(b-a)) times the integral from a to b of f(x) dx.
The definite integral ∫ from a to b of |f(x)| dx represents which geometric quantity?
The total area between the graph of f(x) and the x-axis
The net signed area between the graph of f(x) and the x-axis
The average value of f(x) over [a, b]
The volume of the solid generated by f(x)
By taking the absolute value inside the integral, all segments of the area between the function and the x-axis are treated as positive. This results in the total area, rather than the net (signed) area.
Using implicit differentiation, what is dy/dx for the equation x^2 + y^2 = 25?
-x/y
x/y
-y/x
y/x
Differentiating both sides of the equation x^2 + y^2 = 25 with respect to x gives 2x + 2y(dy/dx) = 0. Solving for dy/dx leads to dy/dx = -x/y, which is the slope of the tangent at any point on the circle.
What is the derivative of f(x) = x^x using logarithmic differentiation?
x^x (ln x + 1)
x^x ln x
x^(x-1) (ln x + 1)
x^x (1 - ln x)
By taking the natural log of f(x) = x^x, we get ln f(x) = x ln x. Differentiating both sides implicitly yields f'(x)/f(x) = ln x + 1, and multiplying by x^x gives f'(x) = x^x (ln x + 1).
Which integral represents the volume of the solid obtained by rotating the region bounded by y = √x and y = 0 about the x-axis from x = 0 to 4?
π ∫ from 0 to 4 of x dx
π ∫ from 0 to 4 of √x dx
2π ∫ from 0 to 4 of x dx
π ∫ from 0 to 4 of x^2 dx
When rotating the area between y = √x and the x-axis about the x-axis, the washer method is applied. Squaring the function gives (√x)^2 = x, so the volume is expressed as π times the integral from 0 to 4 of x dx.
A balloon rises vertically at 5 m/s while a car travels horizontally at 20 m/s. At a moment when the balloon is 30 m above the ground and the car is 40 m away from the point directly below the balloon, what is the rate of change of the distance between them?
19 m/s
15 m/s
20 m/s
25 m/s
Using the Pythagorean theorem, if d^2 = x^2 + y^2, then differentiating gives dd/dt = (x(dx/dt) + y(dy/dt)) / d. Substituting x = 40, dx/dt = 20, y = 30, dy/dt = 5, and d = 50 results in dd/dt = 19 m/s.
Evaluate the limit lim (x→0) (1 - cos(x)) / x^2.
1/2
0
1
2
By either using the Taylor series expansion for cos(x) or L'Hôpital's Rule, one finds that 1 - cos(x) is approximately x^2/2 for small x. Dividing by x^2 gives a limit of 1/2.
Find the critical points for the function f(x) = x^3 - 6x^2 + 9x + 15.
x = 1 and x = 3
x = 2
x = 0 and x = 3
x = -1 and x = 3
Differentiating the function yields f'(x) = 3x^2 - 12x + 9, which factors into 3(x - 1)(x - 3). Setting the derivative equal to zero shows that the critical points are at x = 1 and x = 3.
Which expression correctly represents the definite integral ∫ from a to b of f(x) dx as the limit of a Riemann sum?
lim (n→∞) Σ (i=1 to n) f(a + i*(b-a)/n) * ((b-a)/n)
lim (n→∞) Σ (i=1 to n) f(a) * ((b-a)/n)
lim (n→∞) Σ (i=1 to n) f(b) * ((b-a)/n)
lim (n→∞) Σ (i=1 to n) f(a + i*(b-a)/n) / n
A definite integral is defined as the limit of a Riemann sum. The expression lim (n→∞) Σ f(a + i*(b-a)/n) * ((b-a)/n) is the standard formulation representing ∫ from a to b of f(x) dx.
0
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Study Outcomes

  1. Analyze limits and continuity to determine function behaviors.
  2. Apply differentiation techniques to solve complex calculus problems.
  3. Understand the concept of derivatives and its practical applications.
  4. Synthesize integration methods to compute areas under curves.
  5. Evaluate the relationships between functions and their rates of change.
  6. Interpret real-world scenarios through the lens of calculus problem-solving.

Unit 5 AP Calc AB Review Cheat Sheet

  1. Mean Value Theorem (MVT) - Imagine driving from A to B and knowing your average speed; MVT guarantees there's at least one moment when your instantaneous speed exactly matches that average. This theorem is your gateway to understanding how functions behave between two points. Fiveable MVT Guide
  2. Extreme Value Theorem (EVT) - Think of EVT like picking the tallest and shortest players in a basketball game: if everyone's on the court (continuity), you're guaranteed a highest point and a lowest point. This ensures global peaks and valleys always exist on closed intervals. Fiveable EVT Guide
  3. First Derivative Test - By tracking where f′ changes sign, you can pinpoint where your graph climbs versus where it dives. A switch from positive to negative signals a local maximum (a peak), while negative to positive marks a local minimum (a valley). Meryl's Unit 5 Review
  4. Second Derivative Test - Peek at f″ at your critical points to see if the curve smiles or frowns. If f″(c) > 0, you've got a concave‑up bowl and a local minimum; if f″(c) < 0, it's concave‑down and you've hit a local maximum. Meryl's Unit 5 Review
  5. Concavity & Points of Inflection - Concavity tells you if the curve cups up like a bowl (f″ > 0) or arches down like a bridge (f″ < 0). Points of inflection are the dramatic moments where concavity flips and the graph changes its bend. Meryl's Unit 5 Review
  6. Graphing Functions Using Derivatives - First derivatives reveal where your function is rising or falling, and second derivatives show how sharply it bends. Combining both lets you sketch accurate and detailed graphs without breaking a sweat. Teaching Calculus Unit 5
  7. Connecting a Function & Its Derivatives - Think of f as your road map, f′ as the slope indicator, and f″ as your curvature alert. Mastering their relationship empowers you to forecast every twist and turn of the graph. Teaching Calculus Unit 5
  8. Optimization Problems - Use derivatives to maximize profits, minimize costs, or find ideal dimensions in real‑world scenarios. Set f′ to zero to find critical candidates, then apply derivative tests to crown the winner. Teaching Calculus Unit 5
  9. Understanding Critical Points - Critical points occur where f′ is zero or undefined, acting as hotspots for potential peaks and valleys. Always examine these spots closely to classify whether they're maxima, minima, or flatlanders. Teaching Calculus Unit 5
  10. Applying the Candidates' Test - For closed intervals, evaluate f at each critical point and the endpoints, then compare values to identify the absolute maximum and minimum. This head‑to‑head approach guarantees you find the true global champions. Teaching Calculus Unit 5
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