Which of the following statements is true? Bracketed methods with bounds [math: [a,b]] can only begin if:

Question

BlackTom AI · Accepted Answer

By the Intermediate Value Theorem, if a function f is continuous on [a,b] and f(a) and f(b) have opposite signs (i.e. f(a)·f(b) < 0), then there exists at least one c in (a,b) with f(c) = 0. This sign change of the function values at the endpoints is the basis for bracketing (interval) root-finding methods such as bisection.

1. Relevant knowledge points and concepts
- Continuity on [a,b] and the Intermediate Value Theorem (IVT).
- Bracketed root-finding methods require an interval known to contain a root.
- A sign change at the endpoints (f(a) and f(b) of opposite sign) is the standard condition guaranteeing a root inside the interval.

2. Complete reasoning process step by step
- Start with a continuous f on [a,b].
- Compute f(a) and f(b).
- If f(a)·f(b) < 0, IVT guarantees ∃c ∈ (a,b) s.t. f(c)=0; thus the interval brackets a root and bracketing methods can begin.
- If f(a)·f(b) > 0 (same sign) or equals 0 in a trivial endpoint-root case, the guarantee of an interior root is not given (endpoint could itself be a root if f(a)=0 or f(b)=0).

3. Calculations / intermediate steps (example)
Example: f(x)=x. Take a = −1, b = 2.
f(a) = −1, f(b) = 2, product = −2 < 0 → IVT guarantees a root. Indeed f(0)=0 lies in (−1,2).

4. Why the correct answer is correct
The correct answer states that "Function values at a and b are of opposite signs." That is precisely the f(a)·f(b) < 0 condition required by IVT to ensure a root lies between a and b, hence permitting bracketing methods.

5. Why the other options are incorrect
- a. "a and b are of the same sign": irrelevant — the signs of the endpoint x-values do not determine whether f has a root between them.
- b. "Function values at a and b are of the same sign": incorrect because same-sign function values do not guarantee a root inside (no IVT sign-change guarantee).
- c. "a and b are of opposite signs": also irrelevant — a and b being opposite in sign says nothing about f(a) and f(b); the function values might still be both positive or both negative.

6. Conclusion
Therefore, bracketing methods can only begin if the function values at a and b are of opposite signs, matching the provided correct answer.

Which of the following statements is true? Bracketed methods with bounds [math: [a,b]] can only begin if:单项选择题

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