Limits and Continuity
Evaluate the limit limₓ→2 [(x² – 4)/(x – 2)].
Factor the numerator as (x – 2)(x + 2) and cancel the common factor (x – 2). Then substitute x = 2 to get 2 + 2 = 4.
Differentiate the function (x) = 5x³ – 4x + 7.
Differentiate term by term:
d/dx (5x³) = 15x²
d/dx (–4x) = –4
d/dx (7) = 0
15x² – 4
Find the indefinite integral ∫ 4 dx.
∫ 4 dx = 4x + C, where C is the constant of integration.
Find the critical points of the function f(x) = x³ – 3x² + 2.
For f(x) = x³ – 3x² + 2, f′(x) = 3x² – 6x = 3x(x – 2). Setting f′(x) = 0 gives x = 0 and x = 2.
Verify that the function f(x) = x² – 4x + 3 has an absolute minimum on the interval [0, 5] by finding this minimum value.
For f(x) = x² – 4x + 3, complete the square: x² – 4x + 3 = (x – 2)² – 1. The vertex is at x = 2, yielding a minimum value of –1. (Check endpoints: f(0) = 3 and f(5) = 8.)
Evaluate the limit limₓ→0 [sin(3x)/x].
Rewrite sin(3x)/x as 3 · [sin(3x)/(3x)]. Since limₓ→0 [sin(3x)/(3x)] = 1, the limit equals 3.
Find the derivative of f(x) = cos(x).
The derivative of cos(x) is –sin(x).
Evaluate the definite integral ∫₁³ 2x dx.
∫₁³ 2x dx
= [x²]₁³
= 3² – 1²
= 9 – 1
= 8
Use linear approximation for f(x) = √x at x = 9 to estimate √9.2.
For f(x) = √x, note that f(9) = 3 and f′(x) = 1/(2√x), so f′(9) = 1/6. Thus, using linear approximation:
√(9.2)
≈ f(9) + f′(9)·0.2
≈ 3 + (1/6)(0.2)
≈ 3 + 0.0333.
Given that f(1) = 3 and f(4) = 15, use the Mean Value Theorem to find a value c in (1, 4) such that f′(c) = (f(4) – f(1))/(4 – 1).
The average rate of change over [1, 4] is (15 – 3)/(4 – 1) = 12/3 = 4. By the Mean Value Theorem, there exists c in (1, 4) such that f′(c) = 4.
Use the chain rule to differentiate f(x) = √(2x + 3).
Write f(x) = √(2x + 3)
= (2x + 3)^(1/2).
Then, by the chain rule:
f′(x) = (1/2)(2x + 3)^(–1/2) · 2 = 1/√(2x + 3)
Find the antiderivative of f(x) = 3x² – 6x + 4.
Integrate term by term:
∫ 3x² dx = x³
∫ (–6x) dx = –3x²
∫ 4 dx = 4x
A balloon’s radius is increasing at a constant rate of 0.5 m/s. Find the rate at which the volume is increasing when the radius is 3 m. (Recall: V = (4/3)πr³)
Given V = (4/3)πr³, differentiate with respect to time: dV/dt = 4πr² (dr/dt). When r = 3 m and dr/dt = 0.5 m/s, dV/dt = 4π·9·0.5 = 18π.
Find the value c that satisfies the Mean Value Theorem for f(x) = x² on the interval [2, 5].
For f(x) = x², we have f′(x) = 2x. The average rate of change on [2, 5] is (25 – 4)/(5 – 2) = 21/3 = 7. Setting f′(c) = 7 gives 2c = 7, so c = 3.5.
Use the definition of the derivative to find the derivative of f(x) = x².
Using the definition:
f′(x) = limₕ→0 [(x + h)² – x²] / h
= limₕ→0 [2xh + h²] / h
= limₕ→0 [2x + h] = 2x.
Compute the derivative of f(x) = ln(3x – 2).
For f(x) = ln(3x – 2), apply the chain rule:
f′(x) = 1/(3x – 2) · 3
= 3/(3x – 2)
Evaluate the definite integral ∫₀^(π/2) sin(x) dx.
The antiderivative of sin(x) is –cos(x).
∫₀^(π/2) sin(x) dx
= [–cos(x)]₀^(π/2)
= (–cos(π/2)) + cos(0)
= 0 + 1
= 1
Find the local maximum and minimum values of the function f(x) = x³ – 6x² + 9x + 1 on the interval [0, 4].
For f(x) = x³ – 6x² + 9x + 1, compute f′(x) = 3x² – 12x + 9 = 3(x – 1)(x – 3). The critical points are x = 1 and x = 3. Evaluate:
f(0) = 1
f(1) = 1 – 6 + 9 + 1 = 5
f(3) = 27 – 54 + 27 + 1 = 1
f(4) = 64 – 96 + 36 + 1 = 5
The minimum value is 1 and the maximum value is 5.
Use the Fundamental Theorem of Calculus to evaluate ∫₀¹ (6x² – 4x + 1) dx.
An antiderivative of 6x² – 4x + 1 is 2x³ – 2x² + x. Evaluating from 0 to 1 gives: (2 – 2 + 1) – 0 = 1.
Evaluate the limit limₓ→0 [(e(2x) – e-x/(3x)].
For small x, use the approximations: e^(2x) ≈ 1 + 2x and e^(–x) ≈ 1 – x. Thus, the numerator ≈ (1 + 2x) – (1 – x) = 3x. Dividing by 3x gives 1.
Use implicit differentiation to find dy/dx if x² + y² = 16.
Differentiate x² + y² = 16 implicitly:
2x + 2y (dy/dx) = 0 ⟹ dy/dx = –x/y
Compute the definite integral ∫₁⁴ (1/x) dx.
∫₁⁴ (1/x) dx = ln|x| evaluated from 1 to 4 gives ln(4) – ln(1) = ln(4) – 0 = ln(4).
Determine the volume of the solid obtained by rotating the region bounded by y = x and y = x² about the x-axis from x = 0 to x = 1. (Hint: Use the washer method.)
When rotating the region between y = x (outer radius) and y = x² (inner radius) about the x-axis from x = 0 to 1, use the washer method: V = π ∫₀¹ [(x)² – (x²)²] dx = π ∫₀¹ (x² – x⁴) dx. Evaluate: ∫ x² dx = x³/3 and ∫ x⁴ dx = x⁵/5, so V = π [(1/3) – (1/5)] = π[(5 – 3)/15] = (2π)/15.
Use the Squeeze Theorem to evaluate limₓ→0 [x² · cos(1/x)].
Since –1 ≤ cos(1/x) ≤ 1, multiplying by x² gives –x² ≤ x² cos(1/x) ≤ x². As x → 0, both –x² and x² approach 0. By the Squeeze Theorem, the limit is 0.