Justifying a Solution Exists: The Intermediate Value and Rolle Arguments
A student is analyzing g(x)=x^3-3x+1 on the interval [-2,2]. She computes g(-2)=-1, g(-1)=3, g(1)=-1, g(2)=3. She concludes that g has exactly one root in [-2,2] because the sign changes suggest roots and the graph 'looks like' a cubic. Her partner notes the sign changes occur in three subintervals and asks her to justify the number of roots rigorously rather than by inspection. The teacher asks for a complete argument using the Intermediate Value Theorem to establish existence and Rolle's Theorem (or the derivative) to establish the count, and to explain why 'a cubic can have at most three real roots' needs justification, not assertion.
1. Use the IVT to show g has at least one root in each of the intervals where the sign changes. State the hypotheses you are checking.
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g is a polynomial, hence continuous everywhere, so it is continuous on each closed subinterval. Evaluate: g(-2)=-1<0, g(-1)=3>0, so on [-2,-1] the value 0 lies between g(-2) and g(-1); by IVT there is c1 in (-2,-1) with g(c1)=0. Similarly g(-1)=3>0 and g(1)=-1<0 gives c2 in (-1,1) with g(c2)=0. And g(1)=-1<0, g(2)=3>0 gives c3 in (1,2) with g(c3)=0. So at least three roots exist. The hypotheses used are continuity on the closed interval and the target value 0 lying between the endpoint values.
2. Use Rolle's Theorem or the derivative to show g has at most three real roots on [-2,2].
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Suppose g had four distinct roots a<b<c<d in [-2,2]. Apply Rolle's Theorem on each consecutive pair: there exist points in (a,b),(b,c),(c,d) where g'=0. So g' would have at least three distinct zeros. But g'(x)=3x^2-3=3(x-1)(x+1) has exactly two zeros, x=-1 and x=1. Contradiction. Hence g has at most three roots. Combined with the three found by IVT, g has exactly three roots in [-2,2]. This is the rigorous count: existence from IVT, upper bound from the derivative's zero count via Rolle.
3. The student says 'a cubic has at most three real roots.' Is that a valid justification by itself? How does the argument above supply what the assertion lacks?
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The bare assertion is a true fact but not a justification; it is a remembered theorem without proof in context. The argument above supplies the justification: the derivative of a cubic is quadratic, a quadratic has at most two real zeros, and Rolle's Theorem converts n+1 distinct roots of g into at least n distinct zeros of g'. So a cubic (derivative quadratic, at most two zeros) has at most three roots. The reasoning is the point; citing the fact alone would not demonstrate understanding of why it holds or how Rolle bridges roots of f to zeros of f'.