AP Calc AB Unit 5 Review: Key Concepts and Practice Tips

AP Calculus AB Unit 5 focuses on applying derivatives to analyze functions and solve problems involving real-world rates of change. This unit builds on basic differentiation skills by emphasizing interpretation: how derivatives describe a function’s behavior, including where it increases or decreases, its concavity, and where it reaches maximum or minimum values. Mastery of these ideas supports success in graph analysis, optimization, and motion problems—all key components of the AP exam.

Understanding the First Derivative

The first derivative of a function, $ f’(x) $, represents the instantaneous rate of change of $ f(x) $. When $ f’(x) > 0 $, the function is increasing; when $ f’(x) < 0 $, it is decreasing. Critical points occur where $ f’(x) = 0 $ or $ f’(x) $ is undefined. These points are candidates for relative maxima or minima.

To classify critical points, use the first derivative test. Choose test points to the left and right of each critical number and evaluate the sign of $ f’(x) $. If the sign changes from positive to negative, the function has a relative maximum at that point. A change from negative to positive indicates a relative minimum. This method works even when the second derivative test fails.

For example, consider $ f(x) = x^3 - 6x^2 + 9x $. The derivative is
$$ f’(x) = 3x^2 - 12x + 9. $$
Set $ f’(x) = 0 $:
$$ 3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0. $$
So, $ x = 1 $ and $ x = 3 $ are critical points. Testing intervals:

• For $ x < 1 $, pick $ x = 0 $: $ f’(0) = 9 > 0 $ → increasing
• For $ 1 < x < 3 $, pick $ x = 2 $: $ f’(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 < 0 $ → decreasing
• For $ x > 3 $, pick $ x = 4 $: $ f’(4) = 3(16) - 12(4) + 9 = 48 - 48 + 9 = 9 > 0 $ → increasing

Thus, $ f(x) $ has a relative maximum at $ x = 1 $ and a relative minimum at $ x = 3 $.

Using the Second Derivative

The second derivative, $ f’'(x) $, describes the rate of change of the first derivative. It determines concavity:

• If $ f’'(x) > 0 $, the graph is concave up.
• If $ f’'(x) < 0 $, the graph is concave down.

Points where concavity changes are called inflection points. These occur where $ f’‘(x) = 0 $ or is undefined, and the sign of $ f’'(x) $ changes across that point.

The second derivative test helps classify critical points:

• If $ f’'© > 0 $, then $ f $ has a relative minimum at $ x = c $.
• If $ f’'© < 0 $, then $ f $ has a relative maximum at $ x = c $.
• If $ f’'© = 0 $, the test is inconclusive.

Returning to $ f(x) = x^3 - 6x^2 + 9x $:
$$ f’(x) = 3x^2 - 12x + 9, \quad f’‘(x) = 6x - 12. $$
At $ x = 1 $: $ f’‘(1) = 6(1) - 12 = -6 < 0 $ → relative maximum.
At $ x = 3 $: $ f’'(3) = 6(3) - 12 = 6 > 0 $ → relative minimum.

This confirms the earlier result from the first derivative test.

Optimization Problems

Optimization problems require finding maximum or minimum values of a function under constraints. The process involves:

1. Define variables.
2. Write the function to optimize (e.g., area, cost, volume).
3. Use constraints to eliminate extra variables.
4. Differentiate and find critical points.
5. Use the first or second derivative test to confirm max or min.
6. Check endpoints or domain restrictions.

Example: Maximizing Area

A farmer has 200 meters of fencing and wants to enclose a rectangular field using one side of a barn as part of the boundary. Let $ x $ be the length perpendicular to the barn and $ y $ the length parallel. The total fencing used is $ 2x + y = 200 $, so $ y = 200 - 2x $. The area is:
$$ A = x \cdot y = x(200 - 2x) = 200x - 2x^2. $$
Take the derivative:
$$ A’(x) = 200 - 4x. $$
Set $ A’(x) = 0 $:
$$ 200 - 4x = 0 \Rightarrow x = 50. $$
Now check the second derivative:
$$ A’'(x) = -4 < 0, $$
which confirms a maximum. Then $ y = 200 - 2(50) = 100 $. So the optimal dimensions are $ x = 50 $ meters and $ y = 100 $ meters.

Connecting Graphs and Derivatives

Students must interpret graphs of $ f(x) $, $ f’(x) $, and $ f’'(x) $ together. Key relationships include:

• $ f’(x) > 0 $ → $ f(x) $ increasing
• $ f’(x) < 0 $ → $ f(x) $ decreasing
• $ f’'(x) > 0 $ → $ f(x) $ concave up
• $ f’'(x) < 0 $ → $ f(x) $ concave down

If given a graph of $ f’(x) $, you can determine where $ f(x) $ has maxima, minima, or inflection points. For example, where $ f’(x) $ crosses zero from positive to negative, $ f(x) $ has a relative maximum. Where $ f’'(x) $ changes sign, $ f(x) $ has an inflection point.

These skills are tested in both multiple-choice and free-response questions. Practice identifying features from derivative graphs to strengthen analytical reasoning.

Analyzing Motion Using Derivatives

Unit 5 applies derivatives to motion along a straight line. If $ s(t) $ is the position function, then:

• $ v(t) = s’(t) $ is velocity
• $ a(t) = s’'(t) $ is acceleration

Signs of velocity and acceleration determine motion:

• If $ v(t) > 0 $, object moves right (or up).
• If $ v(t) < 0 $, object moves left (or down).
• If $ v(t) $ and $ a(t) $ have the same sign, the object is speeding up.
• If they have opposite signs, the object is slowing down.

For $ s(t) = t^3 - 6t^2 + 9t $:
$$ v(t) = 3t^2 - 12t + 9, \quad a(t) = 6t - 12. $$
Solve $ v(t) = 0 $:
$$ 3t^2 - 12t + 9 = 0 \Rightarrow t = 1, 3. $$
At $ t = 2 $, $ a(2) = 6(2) - 12 = 0 $, suggesting a possible change in acceleration. Testing intervals shows velocity changes from positive to negative at $ t = 1 $, then back to positive at $ t = 3 $. This means the object changes direction twice.

Key Takeaways

• The first derivative determines where a function is increasing or decreasing and helps locate relative extrema using sign changes.
• The second derivative reveals concavity and inflection points, and can classify extrema when the test is conclusive.
• Optimization problems require setting up equations, applying constraints, and using derivatives to find maxima or minima.
• Graphical analysis connects features of $ f(x) $, $ f’(x) $, and $ f’'(x) $, and is essential for interpreting real-world contexts and solving AP-style questions.
• Motion problems use derivatives to interpret position, velocity, and acceleration, including when an object speeds up or slows down.

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