One mistake students sometimes make is the so-called `proof by example'. The proof by example refers to the situation where the statement is validated for some examples or cases rather than validating for the general case. For the question below, choose the correct solution. Show that for all .单项选择题
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Question

One mistake students sometimes make is the so-called `proof by example'. The proof by example refers to the situation where the statement is validated for some examples or cases rather than validating for the general case. For the question below, choose the correct solution. Show that for all .单项选择题

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BlackTom AI · Accepted Answer

Factor x^2 − 1 as (x + 1)(x − 1). Then analyze the signs of the two factors on the interval.

1. Relevant knowledge points and concepts
- Factoring quadratic expressions: x^2 − 1 = (x + 1)(x − 1).
- Sign analysis on an interval: if A ≥ 0 and B ≤ 0, then A·B ≤ 0.
- Equality cases occur when a factor equals 0.

2. Step-by-step reasoning
- For any x ∈ [−1, 1], compute the two linear factors:
  - x + 1: minimum at x = −1 gives x + 1 = 0, and for larger x this quantity is > 0. Hence x + 1 ≥ 0 on [−1, 1].
  - x − 1: maximum at x = 1 gives x − 1 = 0, and for smaller x this quantity is < 0. Hence x − 1 ≤ 0 on [−1, 1].
- Multiply the inequalities via the factorization:
  x^2 − 1 = (x + 1)(x − 1). Since (x + 1) ≥ 0 and (x − 1) ≤ 0, their product is ≤ 0 for every x in [−1, 1].

3. Intermediate calculations (explicit)
- At x = −1: x^2 − 1 = (−1)^2 − 1 = 0.
- At x = 0: x^2 − 1 = 0 − 1 = −1 < 0.
- At x = 1: x^2 − 1 = 1 − 1 = 0.
- General: (x + 1) ∈ [0, 2], (x − 1) ∈ [−2, 0], so product ∈ [−4, 0], in particular ≤ 0.

4. Why the correct answer is correct
Because the sign of each factor on [−1,1] is determined and their product of a nonnegative and a nonpositive number cannot be positive. This directly yields x^2 − 1 ≤ 0 for all x in the interval, with equality only at x = ±1.

5. Why the other options are incorrect
- Option a (proof by example): checking a few points (e.g. −1, 0, 1, ±0.5) does not prove the inequality for every x in the interval; it is insufficient logically.
- Option c (graph drawing): a graph can illustrate the claim but is not a formal analytical proof in this context; it lacks the explicit sign reasoning or calculus justification needed for a rigorous proof.

6. Conclusion
Therefore, the correct justification is option b: If −1 ≤ x ≤ 1 then x + 1 ≥ 0 and x − 1 ≤ 0, so x^2 − 1 = (x + 1)(x − 1) ≤ 0 for all x ∈ [−1, 1]. This matches the provided answer.