Use the bisection method on f(x) = x² − 7 with the starting interval [2, 3]. Find the midpoint c3 of step 3.
Numerical methods worksheet with answers
1.Easy
f(x) = x² − 7[a, b] = [2, 3]n = 3
c3 =
2.Easy
Apply Newton–Raphson to f(x) = e−x − x with the starting value x₀ = 1. Find x2.
f(x) = e−x − xf′(x) = −e−x − 1x₀ = 1
x2 =
3.Medium
The trapezoid rule is used for ∫01 x³ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 1]. Find the smallest number of strips n that guarantees an error below 1 × 10⁻⁵.
f(x) = x³[a, b] = [0, 1]ε = 1 × 10⁻⁵
n =
4.Medium
Approximate ∫01.5 √(1 + x³) dx with the trapezoid rule using n = 2 strips. Give T.
f(x) = √(1 + x³)[a, b] = [0, 1.5]n = 2
T =
5.Hard
Use Euler's method with h = 0.5 for y′ = e−ty, y(0) = 0.5 to estimate y4 ≈ y(2).
y′ = e−tyy(0) = 0.5h = 0.5
y4 =
6.Hard
Use Heun's method (improved Euler) with h = 0.2 for y′ = e−ty, y(0) = 0 to estimate y2 ≈ y(0.4).
y′ = e−tyy(0) = 0h = 0.2
y2 =
Answer key
- 1. c3 = 2.625
- 2. x2 = 0.567
- 3. n = 224
- 4. T = 2.054
- 5. y4 = 1.497
- 6. y2 = 0.3778
Full step-by-step solutions: practise this topic on examproblems.com.