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The secant method practice problems

Two steps of the secant method from two starting guesses. 30 problems with worked solutions; in 30 of them every equation is proved by a computer algebra system.

5 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 3.635

medium
Use the secant method on \( \displaystyle f(x) = x^{3} - 5 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
  1. x_{n+1} = x_n − f(x_n)(x_n − x_{n−1}) / (f(x_n) − f(x_{n−1})).
    Reviewed
  2. \[ 2 - \frac{\left. x^{3} - 5 \right|_{\substack{ x=2 }}}{- \left. x^{3} - 5 \right|_{\substack{ x=1 }} + \left. x^{3} - 5 \right|_{\substack{ x=2 }}} = \frac{11}{7} \]
    x2 ≈ 1.571429.✓ Proved
  3. \[ \frac{11}{7} + \frac{\left(2 - \frac{11}{7}\right) \left. x^{3} - 5 \right|_{\substack{ x=\frac{11}{7} }}}{- \left. x^{3} - 5 \right|_{\substack{ x=2 }} + \left. x^{3} - 5 \right|_{\substack{ x=\frac{11}{7} }}} = \frac{265}{157} \]
    x3 ≈ 1.687898.✓ Proved
Answer \( x_2 = \frac{11}{7} \approx 1.571429,\ x_3 = \frac{265}{157} \approx 1.687898 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.636

medium
Use the secant method on \( \displaystyle f(x) = x^{2} + 2 x - 4 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
  1. x_{n+1} = x_n − f(x_n)(x_n − x_{n−1}) / (f(x_n) − f(x_{n−1})).
    Reviewed
  2. \[ 2 - \frac{\left. x^{2} + 2 x - 4 \right|_{\substack{ x=2 }}}{- \left. x^{2} + 2 x - 4 \right|_{\substack{ x=1 }} + \left. x^{2} + 2 x - 4 \right|_{\substack{ x=2 }}} = \frac{6}{5} \]
    x2 ≈ 1.200000.✓ Proved
  3. \[ \frac{6}{5} + \frac{\left(2 - \frac{6}{5}\right) \left. x^{2} + 2 x - 4 \right|_{\substack{ x=\frac{6}{5} }}}{- \left. x^{2} + 2 x - 4 \right|_{\substack{ x=2 }} + \left. x^{2} + 2 x - 4 \right|_{\substack{ x=\frac{6}{5} }}} = \frac{16}{13} \]
    x3 ≈ 1.230769.✓ Proved
Answer \( x_2 = \frac{6}{5} \approx 1.200000,\ x_3 = \frac{16}{13} \approx 1.230769 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.640

medium
Use the secant method on \( \displaystyle f(x) = x^{2} - x - 4 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
  1. x_{n+1} = x_n − f(x_n)(x_n − x_{n−1}) / (f(x_n) − f(x_{n−1})).
    Reviewed
  2. \[ 3 - \frac{\left. x^{2} - x - 4 \right|_{\substack{ x=3 }}}{- \left. x^{2} - x - 4 \right|_{\substack{ x=2 }} + \left. x^{2} - x - 4 \right|_{\substack{ x=3 }}} = \frac{5}{2} \]
    x2 ≈ 2.500000.✓ Proved
  3. \[ \frac{5}{2} + \frac{\left(3 - \frac{5}{2}\right) \left. x^{2} - x - 4 \right|_{\substack{ x=\frac{5}{2} }}}{- \left. x^{2} - x - 4 \right|_{\substack{ x=3 }} + \left. x^{2} - x - 4 \right|_{\substack{ x=\frac{5}{2} }}} = \frac{23}{9} \]
    x3 ≈ 2.555556.✓ Proved
Answer \( x_2 = \frac{5}{2} \approx 2.500000,\ x_3 = \frac{23}{9} \approx 2.555556 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.645

medium
Use the secant method on \( \displaystyle f(x) = x^{2} - 2 x - 4 \) with \( \displaystyle x_0 = 3 \) and \( \displaystyle x_1 = 4 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
  1. x_{n+1} = x_n − f(x_n)(x_n − x_{n−1}) / (f(x_n) − f(x_{n−1})).
    Reviewed
  2. \[ 4 - \frac{\left. x^{2} - 2 x - 4 \right|_{\substack{ x=4 }}}{- \left. x^{2} - 2 x - 4 \right|_{\substack{ x=3 }} + \left. x^{2} - 2 x - 4 \right|_{\substack{ x=4 }}} = \frac{16}{5} \]
    x2 ≈ 3.200000.✓ Proved
  3. \[ \frac{16}{5} + \frac{\left(4 - \frac{16}{5}\right) \left. x^{2} - 2 x - 4 \right|_{\substack{ x=\frac{16}{5} }}}{- \left. x^{2} - 2 x - 4 \right|_{\substack{ x=4 }} + \left. x^{2} - 2 x - 4 \right|_{\substack{ x=\frac{16}{5} }}} = \frac{42}{13} \]
    x3 ≈ 3.230769.✓ Proved
Answer \( x_2 = \frac{16}{5} \approx 3.200000,\ x_3 = \frac{42}{13} \approx 3.230769 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 3.649

medium
Use the secant method on \( \displaystyle f(x) = x^{2} + x - 4 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
  1. x_{n+1} = x_n − f(x_n)(x_n − x_{n−1}) / (f(x_n) − f(x_{n−1})).
    Reviewed
  2. \[ 2 - \frac{\left. x^{2} + x - 4 \right|_{\substack{ x=2 }}}{- \left. x^{2} + x - 4 \right|_{\substack{ x=1 }} + \left. x^{2} + x - 4 \right|_{\substack{ x=2 }}} = \frac{3}{2} \]
    x2 ≈ 1.500000.✓ Proved
  3. \[ \frac{3}{2} + \frac{\left(2 - \frac{3}{2}\right) \left. x^{2} + x - 4 \right|_{\substack{ x=\frac{3}{2} }}}{- \left. x^{2} + x - 4 \right|_{\substack{ x=2 }} + \left. x^{2} + x - 4 \right|_{\substack{ x=\frac{3}{2} }}} = \frac{14}{9} \]
    x3 ≈ 1.555556.✓ Proved
Answer \( x_2 = \frac{3}{2} \approx 1.500000,\ x_3 = \frac{14}{9} \approx 1.555556 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More the secant method practice

Use the secant method on \( \displaystyle f(x) = x^{3} - 2 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.637medium✓ Every equation proved
Use the secant method on \( \displaystyle f(x) = x^{2} - 5 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.638medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 3 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.639medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{3} - 3 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.641medium✓ Every equation proved
Use the secant method on \( \displaystyle f(x) = x^{2} + x - 1 \) with \( \displaystyle x_0 = 0 \) and \( \displaystyle x_1 = 1 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.642medium✓ Every equation proved
Use the secant method on \( \displaystyle f(x) = x^{2} - x - 1 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.643medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - x - 3 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.644medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 3 x - 1 \) with \( \displaystyle x_0 = 3 \) and \( \displaystyle x_1 = 4 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.646medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 2 x - 5 \) with \( \displaystyle x_0 = 3 \) and \( \displaystyle x_1 = 4 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.647medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 7 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.648medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 2 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.700medium✓ Every equation proved
Use the secant method on \( \displaystyle f(x) = x^{2} - 3 x - 3 \) with \( \displaystyle x_0 = 3 \) and \( \displaystyle x_1 = 4 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.701medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 3 x - 2 \) with \( \displaystyle x_0 = 3 \) and \( \displaystyle x_1 = 4 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.702medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - x - 5 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.703medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + 2 x - 5 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.704medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 2 x - 1 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.720medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + x - 3 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.721medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + x - 5 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.722medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} - 2 x - 2 \) with \( \displaystyle x_0 = 2 \) and \( \displaystyle x_1 = 3 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.723medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + 3 x - 3 \) with \( \displaystyle x_0 = 0 \) and \( \displaystyle x_1 = 1 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.724medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + 2 x - 2 \) with \( \displaystyle x_0 = 0 \) and \( \displaystyle x_1 = 1 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.725medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + 2 x - 1 \) with \( \displaystyle x_0 = 0 \) and \( \displaystyle x_1 = 1 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.726medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + 3 x - 5 \) with \( \displaystyle x_0 = 1 \) and \( \displaystyle x_1 = 2 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.727medium✓ Nihil obstat
Use the secant method on \( \displaystyle f(x) = x^{2} + 3 x - 2 \) with \( \displaystyle x_0 = 0 \) and \( \displaystyle x_1 = 1 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.728medium✓ Every equation proved
Use the secant method on \( \displaystyle f(x) = x^{2} + 3 x - 1 \) with \( \displaystyle x_0 = 0 \) and \( \displaystyle x_1 = 1 \) to find \( \displaystyle x_2 \) and \( \displaystyle x_3 \).
Problem 3.739medium✓ Nihil obstat