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 − 5x) − x).
Find the limit L = limx→0 (1 − cos 5x)/(4x²), with x in radians.
Let f(x) = x3·e−0.5x. Find f′(1).
The tangent line to f(x) = −x2 + 5 at x = 3 crosses the y-axis at y = b. Find b.
A closed cylindrical can (with a top and a bottom) must hold 330 cm³. The metal used is proportional to the total surface area. Find the height h of the can that uses the least material.
A company sells q units per week. The price at which it can sell them is p = 190 − 0.5q euros per unit, and the cost is C(q) = 5800 + 20q euros. Find the weekly output q that maximises profit.
Evaluate I = ∫01 2x·(x² + 1)1 dx.
Find the area A of the region enclosed by y = x2 and y = 4x.
Answer: L = −2.5
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 = 3.125
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′(1) = 1.516
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 = 14
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: h = 7.49 cm
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.
A″(r) = 4π + 1320/r³ > 0, so this is a minimum.
The cheapest closed can is as tall as it is wide: h = 2r.
Answer: q = 170 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) = −1 < 0, so this is a maximum.
Answer: I = 1.5
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Answer: A = 10.67
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
ExamProblems · examproblems.com/courses/calculus-1