Identifying which theorem statement could be false from a continuous function table

Key concept: This question tests how to match the hypotheses of the Intermediate Value Theorem, Mean Value Theorem, and Extreme Value Theorem to the information actually given.

Step by step

What information the table gives

The table tells us that the function is continuous and records values at four points: $f(2)=1$, $f(3)=14$, $f(4)=20$, and $f(5)=31$. That is enough to use the Intermediate Value Theorem and the Extreme Value Theorem on the interval $[2,5]$, because continuity on a closed interval is the key hypothesis for both.

It is not enough to use the Mean Value Theorem. The Mean Value Theorem requires not only continuity on $[2,5]$ but also differentiability on $(2,5)$. The problem only says the function is continuous, so any statement claiming the MVT must work is the one to scrutinize.

Check each statement carefully

A says IVT gives a value $x$ such that $f(x)=10$. Since $10$ lies between $f(2)=1$ and $f(3)=14$, and $f$ is continuous on $[2,3]$, this must be true.

C and D are both consequences of the Extreme Value Theorem. A continuous function on a closed interval must attain both an absolute minimum and an absolute maximum somewhere on that interval, so these statements are true.

B is the only statement that could be false. The average rate of change from $x=2$ to $x=5$ is

$\frac{f(5)-f(2)}{5-2}=\frac{31-1}{3}=10,$

so if the function were differentiable on $(2,5)$, the MVT would guarantee some $c$ with $f'(c)=10$. But differentiability is not given, so the conclusion is not guaranteed.

Correct choice and why it works

The correct answer is B. The key skill is to verify the hypothesis first:

• IVT needs continuity only.
• EVT needs continuity only.
• MVT needs continuity and differentiability.

Because the problem only states continuity, the MVT statement is the one that could fail.

A useful habit on these questions is to separate what is known from what is merely suggested by the data. A table of values can support an IVT or EVT argument, but it cannot justify differentiability unless the problem explicitly says so.

Pitfall alert

A common mistake is to see the clean table values and assume the function must be smooth enough for the Mean Value Theorem. It is not. Continuity does not imply differentiability, and the MVT can fail even when the secant slope is a neat number like 10. Another trap is to think the IVT needs values at every point in the interval; it only needs continuity and two endpoint values that bracket the target output. For this problem, the incorrect theorem claim is the one that sneaks in an extra hypothesis the prompt never gave.

Try different conditions

If the problem were changed to say that $f$ is continuous on $[2,5]$ and differentiable on $(2,5)$, then statement B would no longer be questionable. In that variant, the average slope is still $10$, so the Mean Value Theorem would guarantee some $c\in(2,5)$ with $f'(c)=10$. If instead the interval were changed to $[2,4]$, then the average slope would become $(20-1)/2=9.5$, and the same logic would apply only if differentiability were also stated. The main idea is that the theorem choice depends on both the interval and the smoothness assumptions.

Further reading

Intermediate Value Theorem, Mean Value Theorem, Extreme Value Theorem