Analytical Applications of Differentiation
Unit 5 of AP Calculus AB, worth 15–18% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Mean Value Theorem, extreme values, increasing and decreasing intervals, first and second derivative tests, concavity, curve sketching, optimisation, implicit relations.
How this unit is tested
Start by sorting every problem into one of two families: statements about a function's derivative that let you conclude something about the function itself (MVT, EVT, first/second derivative tests, concavity), or applied problems where you build an equation and optimize it (optimization, implicit relations). For the first family, always write down the hypotheses of the theorem you plan to use and check them against the given function before doing any algebra — continuity on a closed interval and differentiability on the open interval are not optional extras.
For extrema and monotonicity questions, the working pattern is always: find f', solve f'=0 or find where f' is undefined to get critical numbers, then build a sign chart for f' (or f'' for concavity) by testing one point in each subinterval. Read the sign chart to answer whatever the question actually asks — increasing/decreasing, local max/min, or concave up/down — rather than stopping once you have the critical numbers.
Curve sketching problems are just monotonicity and concavity information combined: report intervals of increase/decrease, classify each critical point, report intervals of concavity, and locate inflection points, then describe the shape in that order.
For optimization, translate the words into an objective function and a constraint, use the constraint to reduce the objective to one variable, restrict the domain to values that make physical sense, then apply a derivative test (checking endpoints if the domain is closed) to justify — not just locate — the max or min.
For implicit relations, differentiate both sides of the given equation with respect to x, treating y as a function of x: every term containing y needs the chain rule (multiply by dy/dx) and every product of x and y needs the product rule. Then solve algebraically for dy/dx; a second differentiation gives d²y/dx².
What you have to know
Mean Value Theorem (MVT)
If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c in (a,b) such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
Rolle's Theorem
If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then there exists at least one c in (a,b) such that $f'(c)=0$. (A special case of the MVT.)
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a,b], then f attains an absolute maximum value and an absolute minimum value on [a,b], each occurring at a critical number or an endpoint.
First Derivative Test
If f' changes sign from positive to negative at a critical number c, f has a local maximum at c; if it changes from negative to positive, f has a local minimum at c; if the sign does not change, c is not a local extremum.
Second Derivative Test
If $f'(c)=0$ and $f''(c)>0$, f has a local minimum at c; if $f''(c)<0$, f has a local maximum at c; if $f''(c)=0$, the test is inconclusive and the first derivative test must be used instead.
Concavity and Inflection Points
f is concave up on an interval where $f''>0$ and concave down where $f''<0$. A point where f'' changes sign (and f is defined) is a point of inflection.
14 practice questions
-
For $f(x)=x^3-x$ on $[0,2]$, find the value of c guaranteed by the Mean Value Theorem.
Show the answer
Answer. c = 2/√3 (≈1.155)
The average rate of change is (f(2)-f(0))/2 = (6-0)/2 = 3. Setting f'(x)=3x²-1 equal to 3 gives x²=4/3, so x=2/√3, which lies in (0,2). -
Explain whether the Mean Value Theorem can be applied to $f(x)=|x-1|$ on $[0,2]$.
Show the answer
Answer. No — MVT cannot be applied.
f is continuous on [0,2] but not differentiable at x=1, which lies inside the open interval (0,2). Since the differentiability hypothesis fails, the theorem's conclusion is not guaranteed. -
Find the absolute maximum and minimum values of $f(x)=x^3-3x^2+2$ on $[-1,3]$.
Show the answer
Answer. Absolute max is 2 (at x=0 and x=3); absolute min is -2 (at x=-1 and x=2).
f'(x)=3x²-6x=3x(x-2), giving critical numbers x=0,2. Evaluating f at critical numbers and endpoints: f(-1)=-2, f(0)=2, f(2)=-2, f(3)=2. The largest value is 2 and the smallest is -2, by the Extreme Value Theorem and candidates test. -
For $f(x)=x^4-4x^3$, find the intervals where f is increasing and decreasing.
Show the answer
Answer. Decreasing on (-∞,3); increasing on (3,∞); x=0 is not an extremum.
f'(x)=4x³-12x²=4x²(x-3). Since x² ≥ 0 always, the sign of f' matches the sign of (x-3): negative for x<3, positive for x>3. At x=0 the sign of f' does not change, so it is not a local extremum despite being a critical number. -
Using the first derivative test, classify the critical numbers of $f(x)=x^4-4x^3$.
Show the answer
Answer. x=0 is neither a max nor min; x=3 is a local minimum.
f'(x)=4x²(x-3) is negative on both sides of x=0 (no sign change), so x=0 is not an extremum. At x=3, f' changes from negative to positive, so x=3 is a local minimum, with f(3)=81-108=-27. -
For $g(x)=x^3-6x^2+9x+1$, use the second derivative test to classify its critical numbers.
Show the answer
Answer. Local maximum at x=1 (value 5); local minimum at x=3 (value 1).
g'(x)=3x²-12x+9=3(x-1)(x-3), giving critical numbers x=1,3. g''(x)=6x-12: g''(1)=-6<0 (local max), g''(3)=6>0 (local min). Substituting gives g(1)=5 and g(3)=1. -
Find the intervals of concavity and any inflection points of $h(x)=x^4-6x^2$.
Show the answer
Answer. Concave down on (-1,1); concave up on (-∞,-1) and (1,∞); inflection points at (-1,-5) and (1,-5).
h''(x)=12x²-12=12(x-1)(x+1), which is zero at x=±1, negative between them, and positive outside. Since h'' changes sign at both x=±1, both are inflection points, with h(±1)=1-6=-5. -
Given that $f'(x)=(x-1)(x+2)$, describe the increasing/decreasing behavior and classify the critical points of f.
Show the answer
Answer. Increasing on (-∞,-2) and (1,∞); decreasing on (-2,1); local max at x=-2, local min at x=1.
f'(x)=(x-1)(x+2) is positive for x<-2, negative between -2 and 1, and positive for x>1. The sign change from positive to negative at x=-2 gives a local max; the change from negative to positive at x=1 gives a local min. -
A farmer has 400 feet of fencing to enclose a rectangular field bordering a straight river (no fence needed along the river). Find the dimensions that maximize the enclosed area, and state the maximum area.
Show the answer
Answer. 100 ft by 200 ft, giving a maximum area of 20,000 square feet.
Let x be the length of each side perpendicular to the river and y the side parallel to it, so 2x+y=400 and A(x)=x(400-2x)=400x-2x². A'(x)=400-4x=0 gives x=100, so y=200. Since A''(x)=-4<0, this critical point is a maximum, giving A=100·200=20,000 sq ft. -
An open-top box is made by cutting squares of side x from the corners of a 12 in by 12 in sheet and folding up the sides. For what value of x is the volume maximized, and what is that maximum volume?
Show the answer
Answer. x=2 inches, giving a maximum volume of 128 cubic inches.
V(x)=x(12-2x)² on 0<x<6. V'(x)=12(x-2)(x-6), which is zero at x=2 (the only critical point in the domain). Testing signs shows V' changes from positive to negative at x=2, confirming a maximum, and V(2)=2(8)²=128. -
For the circle $x^2+y^2=25$, find dy/dx at the point (3,4) and write the equation of the tangent line there.
Show the answer
Answer. dy/dx = -3/4; tangent line: $y-4=-\frac{3}{4}(x-3)$.
Differentiating implicitly: 2x+2y·y'=0, so y'=-x/y. At (3,4), y'=-3/4. Using point-slope form with the point (3,4) and this slope gives the tangent line equation. -
For the curve $xy=4$, find $\frac{d^2y}{dx^2}$ in terms of x and y.
Show the answer
Answer. $\frac{d^2y}{dx^2}=\frac{2y}{x^2}$
Differentiating xy=4 implicitly: x·y'+y=0, so y'=-y/x. Differentiating again: x·y''+y'+y'=0, so y''=-2y'/x. Substituting y'=-y/x gives y''=2y/x². -
Find all points on the curve $x^2+xy+y^2=27$ where the tangent line is horizontal.
Show the answer
Answer. (3,-6) and (-3,6).
Differentiating implicitly: 2x+y+x·y'+2y·y'=0, so y'=-(2x+y)/(x+2y). A horizontal tangent requires the numerator to be zero: 2x+y=0, so y=-2x. Substituting into the original equation gives 3x²=27, so x=±3, giving points (3,-6) and (-3,6). -
Verify that $f(x)=x^3-x$ satisfies the hypotheses of Rolle's Theorem on [-1,1], and find all values of c guaranteed by the theorem.
Show the answer
Answer. f is continuous and differentiable everywhere, f(-1)=f(1)=0, and c = ±1/√3 both satisfy f'(c)=0.
f is a polynomial, so it is continuous and differentiable on all of ℝ. f(-1)=-1+1=0 and f(1)=1-1=0, so f(-1)=f(1). Setting f'(x)=3x²-1=0 gives x=±1/√3, both of which lie in the open interval (-1,1).
What people get wrong
- Applying the Mean Value Theorem or Rolle's Theorem without checking both hypotheses — instead, explicitly verify continuity on the closed interval and differentiability on the open interval (watch for corners or vertical tangents) before invoking the conclusion.
- Treating every critical number as a local extremum — instead, always confirm with a sign change in f' (first derivative test) or the sign of f'' (second derivative test); some critical numbers are neither, like a horizontal inflection point.
- Making sign errors when building a sign chart for f' or f'' — instead, factor the derivative completely, test one value in each subinterval, and track the sign of each factor separately before multiplying them.
- Assuming the second derivative test always works — instead, when $f''(c)=0$, recognize the test is inconclusive and fall back on the first derivative test.
- In optimization problems, forgetting to restrict the domain to physically sensible values or to check endpoints on a closed interval — instead, derive the domain directly from the constraint and evaluate the objective function at both critical points and endpoints when the domain is closed.
- In implicit differentiation, forgetting the chain rule factor of dy/dx on every y-term, or dropping the product rule on mixed terms like xy — instead, differentiate every term with respect to x methodically, term by term, before solving for dy/dx.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.