A student wants to prove that f(x) = x³ − x − 1 has a root in [1, 2]. They compute f(1) = −1 and f(2) = 5, note that f is a polynomial, and conclude by IVT that there is exactly one root in (1, 2). What error, if any, has the student made?
AThe student forgot to verify that f is continuous on [1, 2] before applying IVT
BThe conclusion should say 'at least one root,' not 'exactly one root' — IVT only guarantees existence, not uniqueness
CIVT cannot be applied because x³ − x − 1 is not defined at the endpoints
DNo error — the conclusion is valid as stated
The student correctly identified continuity (polynomial) and the sign change, so the IVT application is valid. The error is in concluding 'exactly one' root. IVT guarantees *at least one* c where f(c) = 0, but the function could cross zero multiple times in the interval. Uniqueness requires additional reasoning (like showing f is strictly monotone on [1, 2]). This is the core limitation of IVT as an existence theorem — it says 'there is,' not 'there is exactly one.'
Question 2 Multiple Choice
Which condition is absolutely required to apply the Intermediate Value Theorem to a function f on an interval [a, b]?
Af must be differentiable on [a, b]
Bf must be continuous on the closed interval [a, b]
Cf must be increasing on [a, b]
Df must have a defined derivative at both endpoints a and b
Continuity on [a, b] is the essential hypothesis. Differentiability (option A) implies continuity but is a stronger condition than IVT requires — many continuous functions are not differentiable. Monotonicity (option C) is irrelevant. IVT fails without continuity: the function f(x) = −1 for x < 0 and f(x) = 1 for x ≥ 0 takes both positive and negative values on [−1, 1] but is discontinuous at 0 and never equals 0.
Question 3 True / False
If a continuous function f satisfies f(a) > 0 and f(b) < 0, the IVT guarantees there is exactly one c in (a, b) where f(c) = 0.
TTrue
FFalse
Answer: False
IVT guarantees *at least one* such c, not exactly one. A function could cross zero three, five, or any odd number of times between a and b while remaining continuous. For example, f(x) = x(x−1)(x−2) on [−1, 3] has f(−1) < 0, f(3) > 0, and three roots in the interval. Uniqueness would require showing f is strictly monotone (e.g., f' > 0 everywhere), which is a separate argument.
Question 4 True / False
The Intermediate Value Theorem can be used to prove that the equation cos(x) = x has a solution, even though there is no algebraic formula for that solution.
TTrue
FFalse
Answer: True
True — this is IVT as an existence theorem. Let g(x) = cos(x) − x. Then g(0) = 1 > 0 and g(π) = −1 − π < 0. Since g is continuous (composition of continuous functions) and changes sign on [0, π], IVT guarantees at least one c ∈ (0, π) where g(c) = 0, i.e., cos(c) = c. The exact value (≈ 0.739) cannot be expressed in closed form, but existence is proven. This illustrates the power of IVT: it establishes mathematical facts that pure algebra cannot.
Question 5 Short Answer
Explain why the IVT is called an 'existence theorem' rather than a 'construction theorem,' and why this distinction matters in practice.
Think about your answer, then reveal below.
Model answer: IVT proves that a value c must exist in an interval but provides no method to find c, no formula for c, and no information about uniqueness — there may be one or many. It is a proof that something is there, not a recipe for locating it. In practice this matters because (1) existence is often the hardest part of a proof — once you know c exists, numerical methods like bisection can approximate it; and (2) it warns you not to ask IVT for more than it delivers. Applying IVT to find 'the' root of x³ = x + 1 still requires a numerical algorithm; IVT only tells you the search is not futile.
The distinction between existence and construction is a recurring theme in mathematics. IVT, the Mean Value Theorem, and the Extreme Value Theorem all guarantee something exists without locating it. This is not a weakness — it is a different kind of mathematical knowledge. Proving existence is logically complete and can justify further computation or theoretical reasoning even when explicit solutions are unavailable.