Difficulty: Intermediate | Prerequisites: Algebra II, precalculus, basic derivative rules
This material covers the single-variable foundations tested on the MAT 211 final: evaluating limits at infinity, applying derivative tests to classify extrema, translating word problems into related-rates expressions, and working with definite integrals (properties, average value, and antiderivatives). These topics recur constantly in the multivariable sections of the course, so gaps here cascade forward. If you are comfortable with basic differentiation rules and the idea of an integral as accumulated change, you are ready for this.
Limits at infinity are decided by the highest-power terms in numerator and denominator. The second derivative test tells you whether a critical point is a local max or min. Definite integrals can be split, reversed, and recombined using additive properties, and the average value of a function on an interval is the integral divided by the interval length.
Limit at infinity
The value a function approaches as x grows without bound. For rational functions, compare the degrees of numerator and denominator.
In simple terms, this means: what does the output settle towards when the input gets huge?
Second derivative test
If f'(c) = 0 and f''(c) > 0, the function has a local minimum at c. If f'(c) = 0 and f''(c) < 0, local maximum.
Think of it as: the second derivative tells you whether the curve is cupping upward (minimum) or downward (maximum) at that critical point.
Critical point
A point where f'(x) = 0 or f'(x) does not exist. These are the only candidates for local extrema.
In simple terms, this means: wherever the tangent line is flat or the derivative breaks, something interesting might be happening.
Definite integral (additive property)
For any a < b < c, the integral from a to c equals the integral from a to b plus the integral from b to c.
Think of it as: you can split a journey into legs and add up the distances.
Average value of a function
The average value of f on [a, b] is (1/(b - a)) times the integral from a to b of f(x) dx.
In simple terms, this means: flatten the curve into a rectangle of the same area, and the height of that rectangle is the average value.
Related rates
A technique for finding how fast one quantity changes by relating it to another quantity whose rate of change is known, using implicit differentiation with respect to time.
Think of it as: two changing quantities are linked by an equation, so their rates of change are linked too.
Antiderivative
A function F(x) whose derivative is f(x). The general antiderivative includes a constant C, which you pin down with an initial condition.
In simple terms, this means: integration undoes differentiation, and the +C accounts for the fact that many functions share the same derivative.
Concavity
A function is concave up where f''(x) > 0 (curve opens upward) and concave down where f''(x) < 0 (curve opens downward).
Think of it as: concave up holds water like a cup; concave down spills it.
The natural log function ln(u) requires u > 0. For ln(x² + y²), the expression x² + y² is always non-negative. It equals zero only at the origin (0, 0), where both x and y are zero. Everywhere else, x² + y² > 0.
The domain is the entire xy-plane except the origin.
Note that the domain is not restricted to the first quadrant. Squaring makes negative values of x and y positive, so the sign of x and y does not matter.
To evaluate the limit of a rational function as x approaches infinity, identify the highest power of x in the numerator and denominator.
Same degree in numerator and denominator: the limit is the ratio of the leading coefficients.
Higher degree in the numerator: the limit is ±infinity.
Higher degree in the denominator: the limit is 0.
Worked example from the exam:
lim (x→∞) of (10x² + 10x³ + 1) / (4 + 10x + x² - 5x³)
The highest power in both numerator and denominator is x³. The leading coefficient in the numerator is 10; in the denominator it is -5.
Limit = 10 / (-5) = -2.
Answer: C.
If f'(c) = 0 (critical point) and f''(c) > 0, the function has a local minimum at x = c. If f''(c) < 0, it has a local maximum. If f''(c) = 0, the test is inconclusive.
Worked example from the exam:
Given f'(4) = 0 and f''(4) > 0, what can we conclude? Since f''(4) > 0, the curve is concave up at x = 4. A critical point where the curve is concave up is a local minimum.
Answer: B (local minimum at x = 4).
To determine concavity at a specific point, compute f''(x) and evaluate it there.
True/False example: Is f(x) = x³ - 2x² concave down at x = 2?
f'(x) = 3x² - 4x, so f''(x) = 6x - 4. At x = 2, f''(2) = 12 - 4 = 8 > 0, which means concave up, not concave down.
Answer: False.
A critical point occurs where f'(x) = 0 or where f'(x) does not exist, provided x is in the domain of f.
True/False example: Does f(x) = x^(1/2) have a critical point at x = 0?
f'(x) = (1/2)x^(-1/2) = 1/(2√x). At x = 0, f'(0) is undefined. However, x = 0 is an endpoint of the domain [0, ∞), not an interior point. Whether this counts as a critical point depends on convention. In many textbook definitions, critical points must be interior to the domain, which would make this false. In other conventions where endpoints with undefined derivatives count, it would be true.
The exam expects: True (the derivative does not exist at x = 0, and x = 0 is in the domain).
A function f(x) is increasing at x = a when f'(a) > 0.
Worked example from the exam:
For f(x) = 4x² + qx + 6 to be increasing at x = 1, we need f'(1) > 0.
f'(x) = 8x + q, so f'(1) = 8 + q > 0, which gives q > -8.
The interval for q is (-8, ∞).
Answer: C.
In related rates, identify what is given and what is asked:
"The radius is currently 20 cm" tells you r = 20 (a value, not a rate).
"Increasing by 1 cm per second" tells you dr/dt = 1 cm/sec (a rate).
"How fast is the volume increasing?" asks for dV/dt.
So the correct translation is: given dr/dt = 1 cm/sec, find dV/dt when r = 20 cm.
Answer: D.
True/False example: If y is a function of x and f = xy, is df/dx = y?
No. By the product rule, df/dx = y + x(dy/dx). The term x(dy/dx) cannot be ignored when y depends on x.
Answer: False.
The integral from a to c can be split at any intermediate point b:
∫(a to c) f(x) dx = ∫(a to b) f(x) dx + ∫(b to c) f(x) dx
This works in reverse too. To find a missing piece, rearrange.
Worked example from the exam:
Given ∫(1 to 10) f(x) dx = 3 and ∫(5 to 10) f(x) dx = 7, find ∫(1 to 5) f(x) dx.
∫(1 to 10) = ∫(1 to 5) + ∫(5 to 10), so 3 = ∫(1 to 5) + 7, giving ∫(1 to 5) = -4.
Answer: B.
The average value of f on [a, b] is:
avg = (1/(b - a)) × ∫(a to b) f(x) dx
Worked example from the exam:
For f(t) = e^(t/10) / 40 on [1, 9], the average value integral is:
(1/(9 - 1)) × ∫(1 to 9) [e^(t/10) / 40] dt = (1/8) × ∫(1 to 9) [e^(t/10) / 40] dt
Answer: B.
To recover a total function from a marginal (derivative) function, integrate and use an initial condition to solve for C.
Worked example from the exam:
Given R'(x) = x³ - 20x + 200 and R(3) = 800.
(a) The integral setup: R(x) = ∫ (x³ - 20x + 200) dx
(b) Evaluate the integral:
R(x) = x⁴/4 - 10x² + 200x + C
Use R(3) = 800:
R(3) = 81/4 - 90 + 600 + C = 81/4 + 510 + C = 2121/4 + C = 800
C = 800 - 2121/4 = 3200/4 - 2121/4 = 1079/4
R(x) = x⁴/4 - 10x² + 200x + 1079/4
True/False example: Is ∫(1/x²) dx = ln|x²| + C?
No. The integral of 1/x² = x^(-2) is x^(-1)/(-1) = -1/x + C.
The ln rule applies to ∫(1/x) dx, not ∫(1/x²) dx. The exponent matters.
Answer: False.
Limit of a rational function at infinity: compare leading-term coefficients when degrees match. Limit = (leading coeff. of numerator) / (leading coeff. of denominator).
Second derivative test: f'(c) = 0 and f''(c) > 0 → local min; f''(c) < 0 → local max; f''(c) = 0 → inconclusive.
Average value: avg = [1/(b - a)] × ∫(a to b) f(x) dx
Power rule for integration: ∫ x^n dx = x^(n+1) / (n+1) + C, provided n ≠ -1.
ln rule for integration: ∫ (1/x) dx = ln|x| + C. This applies only when the exponent is exactly -1.
Volume of a sphere: V = (4/3)πr³, so dV/dr = 4πr². For related rates: dV/dt = 4πr² × (dr/dt).
Product rule (when y depends on x): d/dx [xy] = y + x(dy/dx), not simply y.
Students often confuse ∫(1/x²) dx with the ln rule. The ln rule only applies to ∫(1/x) dx. The integral of 1/x² uses the power rule: ∫ x^(-2) dx = -x^(-1) + C = -1/x + C.
Students often forget the product rule when differentiating xy where y depends on x. d/dx(xy) = y + x(dy/dx), not just y. Treating y as a constant is only valid when y is independent of x.
In related rates, students mix up which quantities are rates (dr/dt) and which are values (r = 20). The phrase "increasing by 1 cm per second" is a rate. The phrase "the radius is 20 cm" is a value at a specific moment.
When finding limits at infinity, students sometimes look at the lowest-power terms or the constant terms. Only the highest-power terms matter, because they dominate as x grows.
⚠️ The additive property of integrals appears as a "work backwards" problem. You will be given two of the three integrals and asked to find the third. Be comfortable rearranging.
⚠️ The average value formula requires dividing by (b - a), not by b or by a. On [1, 9], the divisor is 8, not 9.
⚠️ Related rates questions on this exam test your ability to translate English into mathematical notation, not to solve the equation. Read carefully: which quantity is a rate? Which is a snapshot value?
⚠️ Antiderivative problems with initial conditions: integrate first, then plug in the known point to solve for C. Do not forget the constant.
⚠️ True/false questions on this exam are worth 2 points each with no partial credit. Read them slowly. Several are designed to catch common errors (like applying the ln rule to the wrong integrand).
True or false: The limit of (3x⁴ + x) / (7x⁴ - 2) as x → ∞ is 3/7. (True)
Fill in the blank: If f'(5) = 0 and f''(5) = -3, then f has a local ______ at x = 5. (maximum)
True or false: ∫(1/x³) dx = ln|x³| + C. (False; it equals -1/(2x²) + C)
Fill in the blank: The average value of f on [2, 10] uses a multiplier of 1/______. (8)
True or false: In the product rule, d/dx(xy) = y when y depends on x. (False; it equals y + x·dy/dx)
Q: Evaluate lim (x→∞) of (5x³ + 2x) / (x² - 3x³). What is the answer?
A: The highest power in both is x³. Leading coefficients: 5 in the numerator, -3 in the denominator. The limit is 5/(-3) = -5/3.
Q: Given ∫(2 to 8) f(x) dx = 10 and ∫(2 to 5) f(x) dx = 4, find ∫(5 to 8) f(x) dx.
A: By the additive property, ∫(2 to 8) = ∫(2 to 5) + ∫(5 to 8). So 10 = 4 + ∫(5 to 8), giving ∫(5 to 8) = 6.
Q: The marginal cost is C'(x) = 6x² - 4x + 10 and C(0) = 500. Find C(x).
A: C(x) = ∫(6x² - 4x + 10) dx = 2x³ - 2x² + 10x + K. C(0) = 500 → K = 500. So C(x) = 2x³ - 2x² + 10x + 500.
Q: A spherical balloon is inflating so that dr/dt = 2 cm/s. How fast is the volume changing when r = 5 cm?
A: V = (4/3)πr³, so dV/dt = 4πr²(dr/dt) = 4π(25)(2) = 200π cm³/s.
Q: For f(x) = 4x² + qx + 6, what values of q make f increasing at x = 3?
A: f'(x) = 8x + q. f'(3) = 24 + q > 0, so q > -24. The interval is (-24, ∞).
The second derivative test here is the single-variable version of the Hessian test you will use for multivariable functions (see the Multivariable Optimization study notes). The idea is the same: check the sign of curvature at a critical point to classify it.
Related rates connect to partial derivatives: in multivariable calculus, when several variables change with time, the chain rule generalises to dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt). The balloon problem is a one-variable preview of that.
The antiderivative/initial-condition technique reappears whenever you recover a total function from a marginal function in economics: marginal cost, marginal revenue, or marginal profit.
limits at infinity, horizontal asymptote, leading coefficient ratio, rational function limit, second derivative test, concavity, concave up, concave down, inflection point, critical point, derivative equals zero, undefined derivative, average value of a function, mean value theorem for integrals, definite integral properties, additive property of integrals, splitting integrals, related rates, implicit differentiation with time, chain rule, dV/dt, dr/dt, antiderivative, indefinite integral, initial condition, constant of integration, power rule for integration, ln integration rule, product rule with dependent variable, MAT 211, brief calculus, ASU calculus