Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
Find the limit L = limx→∞ (√(x2 + 9x) − x).
Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.
Let f(x) = (4x + 5)3. Find f′(0).
The tangent line to f(x) = 0.5x2 − x + 5 at x = 1 crosses the y-axis at y = b. Find b.
An open box (no lid) with a square base must hold 62500 cm³. Use the least possible material, that is the smallest area of the base plus the four sides. Find the minimum surface area S.
A company sells q units per week. The price at which it can sell them is p = 190 − 2q euros per unit, and the cost is C(q) = 1000 + 75q euros. Find the weekly output q that maximises profit.
Evaluate I = ∫₀π/3 x·sin x dx (x in radians).
The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
Answer: L = 4.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Answer: L = 2.667
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Answer: f′(0) = 300
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Answer: b = 4.5
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Answer: S = 7500 cm²
Optimization: write the quantity to optimise as a function of one variable, using the constraint to eliminate the other. Set the derivative to zero and solve. Check that the critical point is a maximum or minimum (second derivative or sign change of the first derivative) and that it lies inside the allowed range; an endpoint such as a zero side length is not the answer.
S″(x) = 2 + 500000/x³ > 0, so this is a minimum.
The best open box is half as tall as it is wide.
Answer: q = 28.75 units
Optimization: write the quantity to optimise as a function of one variable, using the constraint to eliminate the other. Set the derivative to zero and solve. Check that the critical point is a maximum or minimum (second derivative or sign change of the first derivative) and that it lies inside the allowed range; an endpoint such as a zero side length is not the answer.
P″(q) = −4 < 0, so this is a maximum.
Answer: I = 0.3424
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Answer: V = 3.351
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
ExamProblems · examproblems.com/courses/calculus-1