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Intermediate Value Theorem: locating a root practice problems

Use the Intermediate Value Theorem to show a root exists in an interval, then narrow it by bisection. 58 problems with worked solutions; in 58 of them every equation is proved by a computer algebra system.

16 worked problems

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

Problem 1.607

medium
Show that \( \displaystyle x + e^{x} - 5 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [1, 2] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x + e^{x} - 5 \right|_{\substack{ x=1 }} = -4 + e \]
    f(1).✓ Proved
  3. \[ \left. x + e^{x} - 5 \right|_{\substack{ x=2 }} = -3 + e^{2} \]
    f(2): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x + e^{x} - 5 \right|_{\substack{ x=\frac{3}{2} }} = - \frac{7}{2} + e^{\frac{3}{2}} \]
    f(3/2) ≈ 0.9817.✓ Proved
  5. \[ \left. x + e^{x} - 5 \right|_{\substack{ x=\frac{5}{4} }} = - \frac{15}{4} + e^{\frac{5}{4}} \]
    f(5/4) ≈ -0.2597.✓ Proved
  6. The sign changes on [5/4, 3/2], so a root lies there.
    Reviewed
Answer \( \left[\frac{5}{4}, \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.608

medium
Show that \( \displaystyle - x + \cos{\left(x \right)} = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [0, 1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. - x + \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]
    f(0).✓ Proved
  3. \[ \left. - x + \cos{\left(x \right)} \right|_{\substack{ x=1 }} = -1 + \cos{\left(1 \right)} \]
    f(1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. - x + \cos{\left(x \right)} \right|_{\substack{ x=1 \cdot \frac{1}{2} }} = - \frac{1}{2} + \cos{\left(\frac{1}{2} \right)} \]
    f(1/2) ≈ 0.3776.✓ Proved
  5. \[ \left. - x + \cos{\left(x \right)} \right|_{\substack{ x=\frac{3}{4} }} = - \frac{3}{4} + \cos{\left(\frac{3}{4} \right)} \]
    f(3/4) ≈ -0.0183.✓ Proved
  6. The sign changes on [1/2, 3/4], so a root lies there.
    Reviewed
Answer \( \left[\frac{1}{2}, \frac{3}{4}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.610

medium
Show that \( \displaystyle x^{5} + x^{2} - 5 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [1, 2] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{5} + x^{2} - 5 \right|_{\substack{ x=1 }} = -3 \]
    f(1).✓ Proved
  3. \[ \left. x^{5} + x^{2} - 5 \right|_{\substack{ x=2 }} = 31 \]
    f(2): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{5} + x^{2} - 5 \right|_{\substack{ x=\frac{3}{2} }} = \frac{155}{32} \]
    f(3/2) ≈ 4.8438.✓ Proved
  5. \[ \left. x^{5} + x^{2} - 5 \right|_{\substack{ x=\frac{5}{4} }} = - \frac{395}{1024} \]
    f(5/4) ≈ -0.3857.✓ Proved
  6. The sign changes on [5/4, 3/2], so a root lies there.
    Reviewed
Answer \( \left[\frac{5}{4}, \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.615

medium
Show that \( \displaystyle x^{3} - 3 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [1, 2] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - 3 \right|_{\substack{ x=1 }} = -2 \]
    f(1).✓ Proved
  3. \[ \left. x^{3} - 3 \right|_{\substack{ x=2 }} = 5 \]
    f(2): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - 3 \right|_{\substack{ x=\frac{3}{2} }} = \frac{3}{8} \]
    f(3/2) ≈ 0.3750.✓ Proved
  5. \[ \left. x^{3} - 3 \right|_{\substack{ x=\frac{5}{4} }} = - \frac{67}{64} \]
    f(5/4) ≈ -1.0469.✓ Proved
  6. The sign changes on [5/4, 3/2], so a root lies there.
    Reviewed
Answer \( \left[\frac{5}{4}, \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.618

medium
Show that \( \displaystyle x^{3} - 3 x - 1 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - 3 x - 1 \right|_{\substack{ x=-2 }} = -3 \]
    f(-2).✓ Proved
  3. \[ \left. x^{3} - 3 x - 1 \right|_{\substack{ x=-1 }} = 1 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - 3 x - 1 \right|_{\substack{ x=- \frac{3}{2} }} = \frac{1}{8} \]
    f(-3/2) ≈ 0.1250.✓ Proved
  5. \[ \left. x^{3} - 3 x - 1 \right|_{\substack{ x=- \frac{7}{4} }} = - \frac{71}{64} \]
    f(-7/4) ≈ -1.1094.✓ Proved
  6. The sign changes on [-7/4, -3/2], so a root lies there.
    Reviewed
Answer \( \left[- \frac{7}{4}, - \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.638

medium
Show that \( \displaystyle x^{4} + x - 5 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{4} + x - 5 \right|_{\substack{ x=-2 }} = 9 \]
    f(-2).✓ Proved
  3. \[ \left. x^{4} + x - 5 \right|_{\substack{ x=-1 }} = -5 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{4} + x - 5 \right|_{\substack{ x=- \frac{3}{2} }} = - \frac{23}{16} \]
    f(-3/2) ≈ -1.4375.✓ Proved
  5. \[ \left. x^{4} + x - 5 \right|_{\substack{ x=- \frac{7}{4} }} = \frac{673}{256} \]
    f(-7/4) ≈ 2.6289.✓ Proved
  6. The sign changes on [-7/4, -3/2], so a root lies there.
    Reviewed
Answer \( \left[- \frac{7}{4}, - \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.644

medium
Show that \( \displaystyle x^{3} - 3 x = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - 3 x \right|_{\substack{ x=-2 }} = -2 \]
    f(-2).✓ Proved
  3. \[ \left. x^{3} - 3 x \right|_{\substack{ x=-1 }} = 2 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - 3 x \right|_{\substack{ x=- \frac{3}{2} }} = \frac{9}{8} \]
    f(-3/2) ≈ 1.1250.✓ Proved
  5. \[ \left. x^{3} - 3 x \right|_{\substack{ x=- \frac{7}{4} }} = - \frac{7}{64} \]
    f(-7/4) ≈ -0.1094.✓ Proved
  6. The sign changes on [-7/4, -3/2], so a root lies there.
    Reviewed
Answer \( \left[- \frac{7}{4}, - \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.677

medium
Show that \( \displaystyle x^{3} + 2 x - 1 = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [0, 1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} + 2 x - 1 \right|_{\substack{ x=0 }} = -1 \]
    f(0).✓ Proved
  3. \[ \left. x^{3} + 2 x - 1 \right|_{\substack{ x=1 }} = 2 \]
    f(1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} + 2 x - 1 \right|_{\substack{ x=1 \cdot \frac{1}{2} }} = \frac{1}{8} \]
    f(1/2) ≈ 0.1250.✓ Proved
  5. \[ \left. x^{3} + 2 x - 1 \right|_{\substack{ x=1 \cdot \frac{1}{4} }} = - \frac{31}{64} \]
    f(1/4) ≈ -0.4844.✓ Proved
  6. The sign changes on [1/4, 1/2], so a root lies there.
    Reviewed
Answer \( \left[\frac{1}{4}, \frac{1}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.678

medium
Show that \( \displaystyle x^{3} - x - 3 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [1, 2] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - x - 3 \right|_{\substack{ x=1 }} = -3 \]
    f(1).✓ Proved
  3. \[ \left. x^{3} - x - 3 \right|_{\substack{ x=2 }} = 3 \]
    f(2): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - x - 3 \right|_{\substack{ x=\frac{3}{2} }} = - \frac{9}{8} \]
    f(3/2) ≈ -1.1250.✓ Proved
  5. \[ \left. x^{3} - x - 3 \right|_{\substack{ x=\frac{7}{4} }} = \frac{39}{64} \]
    f(7/4) ≈ 0.6094.✓ Proved
  6. The sign changes on [3/2, 7/4], so a root lies there.
    Reviewed
Answer \( \left[\frac{3}{2}, \frac{7}{4}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.681

medium
Show that \( \displaystyle x^{3} + x - 1 = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [0, 1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} + x - 1 \right|_{\substack{ x=0 }} = -1 \]
    f(0).✓ Proved
  3. \[ \left. x^{3} + x - 1 \right|_{\substack{ x=1 }} = 1 \]
    f(1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} + x - 1 \right|_{\substack{ x=1 \cdot \frac{1}{2} }} = - \frac{3}{8} \]
    f(1/2) ≈ -0.3750.✓ Proved
  5. \[ \left. x^{3} + x - 1 \right|_{\substack{ x=\frac{3}{4} }} = \frac{11}{64} \]
    f(3/4) ≈ 0.1719.✓ Proved
  6. The sign changes on [1/2, 3/4], so a root lies there.
    Reviewed
Answer \( \left[\frac{1}{2}, \frac{3}{4}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.685

medium
Show that \( \displaystyle x^{3} - x + 2 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - x + 2 \right|_{\substack{ x=-2 }} = -4 \]
    f(-2).✓ Proved
  3. \[ \left. x^{3} - x + 2 \right|_{\substack{ x=-1 }} = 2 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - x + 2 \right|_{\substack{ x=- \frac{3}{2} }} = \frac{1}{8} \]
    f(-3/2) ≈ 0.1250.✓ Proved
  5. \[ \left. x^{3} - x + 2 \right|_{\substack{ x=- \frac{7}{4} }} = - \frac{103}{64} \]
    f(-7/4) ≈ -1.6094.✓ Proved
  6. The sign changes on [-7/4, -3/2], so a root lies there.
    Reviewed
Answer \( \left[- \frac{7}{4}, - \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.716

medium
Show that \( \displaystyle x^{3} + 3 x + 3 = 0 \) has a solution in \( \displaystyle [-1, 0] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-1, 0] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} + 3 x + 3 \right|_{\substack{ x=-1 }} = -1 \]
    f(-1).✓ Proved
  3. \[ \left. x^{3} + 3 x + 3 \right|_{\substack{ x=0 }} = 3 \]
    f(0): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} + 3 x + 3 \right|_{\substack{ x=- \frac{1}{2} }} = \frac{11}{8} \]
    f(-1/2) ≈ 1.3750.✓ Proved
  5. \[ \left. x^{3} + 3 x + 3 \right|_{\substack{ x=- \frac{3}{4} }} = \frac{21}{64} \]
    f(-3/4) ≈ 0.3281.✓ Proved
  6. The sign changes on [-1, -3/4], so a root lies there.
    Reviewed
Answer \( \left[-1, - \frac{3}{4}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.745

medium
Show that \( \displaystyle x^{3} - 2 x + 1 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - 2 x + 1 \right|_{\substack{ x=-2 }} = -3 \]
    f(-2).✓ Proved
  3. \[ \left. x^{3} - 2 x + 1 \right|_{\substack{ x=-1 }} = 2 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - 2 x + 1 \right|_{\substack{ x=- \frac{3}{2} }} = \frac{5}{8} \]
    f(-3/2) ≈ 0.6250.✓ Proved
  5. \[ \left. x^{3} - 2 x + 1 \right|_{\substack{ x=- \frac{7}{4} }} = - \frac{55}{64} \]
    f(-7/4) ≈ -0.8594.✓ Proved
  6. The sign changes on [-7/4, -3/2], so a root lies there.
    Reviewed
Answer \( \left[- \frac{7}{4}, - \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.746

medium
Show that \( \displaystyle x^{3} + 3 x + 6 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} + 3 x + 6 \right|_{\substack{ x=-2 }} = -8 \]
    f(-2).✓ Proved
  3. \[ \left. x^{3} + 3 x + 6 \right|_{\substack{ x=-1 }} = 2 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} + 3 x + 6 \right|_{\substack{ x=- \frac{3}{2} }} = - \frac{15}{8} \]
    f(-3/2) ≈ -1.8750.✓ Proved
  5. \[ \left. x^{3} + 3 x + 6 \right|_{\substack{ x=- \frac{5}{4} }} = \frac{19}{64} \]
    f(-5/4) ≈ 0.2969.✓ Proved
  6. The sign changes on [-3/2, -5/4], so a root lies there.
    Reviewed
Answer \( \left[- \frac{3}{2}, - \frac{5}{4}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.749

medium
Show that \( \displaystyle x^{3} + x + 5 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} + x + 5 \right|_{\substack{ x=-2 }} = -5 \]
    f(-2).✓ Proved
  3. \[ \left. x^{3} + x + 5 \right|_{\substack{ x=-1 }} = 3 \]
    f(-1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} + x + 5 \right|_{\substack{ x=- \frac{3}{2} }} = \frac{1}{8} \]
    f(-3/2) ≈ 0.1250.✓ Proved
  5. \[ \left. x^{3} + x + 5 \right|_{\substack{ x=- \frac{7}{4} }} = - \frac{135}{64} \]
    f(-7/4) ≈ -2.1094.✓ Proved
  6. The sign changes on [-7/4, -3/2], so a root lies there.
    Reviewed
Answer \( \left[- \frac{7}{4}, - \frac{3}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 1.823

medium
Show that \( \displaystyle x^{3} - 3 x - 6 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [2, 3] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{3} - 3 x - 6 \right|_{\substack{ x=2 }} = -4 \]
    f(2).✓ Proved
  3. \[ \left. x^{3} - 3 x - 6 \right|_{\substack{ x=3 }} = 12 \]
    f(3): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{3} - 3 x - 6 \right|_{\substack{ x=\frac{5}{2} }} = \frac{17}{8} \]
    f(5/2) ≈ 2.1250.✓ Proved
  5. \[ \left. x^{3} - 3 x - 6 \right|_{\substack{ x=\frac{9}{4} }} = - \frac{87}{64} \]
    f(9/4) ≈ -1.3594.✓ Proved
  6. The sign changes on [9/4, 5/2], so a root lies there.
    Reviewed
Answer \( \left[\frac{9}{4}, \frac{5}{2}\right] \)

✓ Nihil obstat Lines: 4 proved, 2 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 intermediate value theorem: locating a root practice

Show that \( \displaystyle x^{3} - 7 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.605medium✓ Every equation proved
Show that \( \displaystyle x^{5} + x^{2} - 4 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.606medium✓ Every equation proved
Show that \( \displaystyle x + e^{x} - 4 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.609medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 3 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.611medium✓ Every equation proved
Show that \( \displaystyle x + e^{x} - 2 = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.612medium✓ Nihil obstat
Show that \( \displaystyle x + e^{x} - 8 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.613medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 4 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.614medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 9 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.616medium✓ Every equation proved
Show that \( \displaystyle x + e^{x} - 6 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.617medium✓ Nihil obstat
Show that \( \displaystyle x^{5} + x^{2} - 1 = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.619medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 2 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.639medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 6 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.640medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 2 x - 5 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.641medium✓ Every equation proved
Show that \( \displaystyle x^{3} - x + 1 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.642medium✓ Every equation proved
Show that \( \displaystyle x^{3} + x - 3 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.643medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 9 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.645medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 8 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.646medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 4 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.647medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 2 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.673medium✓ Every equation proved
Show that \( \displaystyle x^{5} + x^{2} - 3 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.674medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 5 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.675medium✓ Nihil obstat
Show that \( \displaystyle x^{5} + x^{2} - 6 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.676medium✓ Every equation proved
Show that \( \displaystyle x + e^{x} - 3 = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.679medium✓ Every equation proved
Show that \( \displaystyle x + e^{x} - 7 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.680medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 2 x - 2 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.682medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 2 x - 3 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.683medium✓ Every equation proved
Show that \( \displaystyle x^{3} - x + 4 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.684medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 7 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.686medium✓ Every equation proved
Show that \( \displaystyle x^{4} + x - 6 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.687medium✓ Every equation proved
Show that \( \displaystyle x^{3} + 2 x - 5 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.715medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 2 x = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.741medium✓ Every equation proved
Show that \( \displaystyle x^{3} - x - 2 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.742medium✓ Every equation proved
Show that \( \displaystyle x^{3} + 3 x - 6 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.743medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 3 x - 5 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.744medium✓ Every equation proved
Show that \( \displaystyle x^{3} + 3 x + 2 = 0 \) has a solution in \( \displaystyle [-1, 0] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.747medium✓ Every equation proved
Show that \( \displaystyle x^{3} + 2 x + 1 = 0 \) has a solution in \( \displaystyle [-1, 0] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.748medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 3 x + 1 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.824medium✓ Every equation proved
Show that \( \displaystyle x^{3} - x - 1 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.825medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - 3 x + 6 = 0 \) has a solution in \( \displaystyle [-3, -2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.826medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 2 x - 6 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.827medium✓ Nihil obstat
Show that \( \displaystyle x^{3} - x + 5 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.828medium✓ Every equation proved
Show that \( \displaystyle x^{3} - 3 x - 4 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
Problem 1.829medium✓ Nihil obstat