- Topic
- Continuity
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Continuity practice question 20
The equation has a solution between and . Starting from the interval , take two steps of the bisection method: each time, evaluate at the midpoint and keep the half on which changes sign. Find the midpoint of the interval left after the two steps.
First check that the theorem applies: is a polynomial, so it is continuous, and . Now evaluate at the midpoint .
Keep the half whose endpoints give opposite signs, take its midpoint, and repeat once more.
An exact fraction, or a decimal to three places.
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
Bisection is the Intermediate Value Theorem used again and again: g is continuous, so it has a zero on any interval where its sign changes. Halve the interval, keep the half with the sign change, and repeat.
Working
g is a polynomial, so it is continuous on [0, 1], and by the Intermediate Value Theorem it has a zero in (0, 1).
g changes sign between 0.5 and 1.
Now the sign change is between 0.5 and 0.75.
The midpoint of the interval left after two steps.
Answer
The answer is 0.625, which is five eighths.
Check
The solution is about 0.6823, which lies in [0.5, 0.75], within 0.06 of the estimate.
Where this goes wrong
Keeping the wrong half, or reporting the last point where g was evaluated. The half to keep is the one whose endpoints give g opposite signs, and the estimate is the midpoint of what is left.
Leans on
Evaluating a polynomial, from Power functions and first sketches
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