Applying the Mean Value Theorem on intervals with a rational function

Key takeaway: This question is about verifying the two MVT hypotheses: continuity on the closed interval and differentiability on the open interval.

Step 1: locate the discontinuities

We are given

$f(x)=\frac{1}{(x-4)(x+2)}.$

A rational function is continuous and differentiable everywhere it is defined, so the only places we must worry about are where the denominator is zero. Those points are

$x=4 \quad \text{and} \quad x=-2.$

So any interval containing either $-2$ or $4$ cannot satisfy the continuity requirement for the Mean Value Theorem.

Step 2: test each interval

I. $[-1,3]$

This interval does not include $-2$ or $4$. Therefore $f$ is continuous on $[-1,3]$ and differentiable on $(-1,3)$. The MVT can be applied.

II. $[5,7]$

This interval is entirely to the right of $4$, so it also avoids both discontinuities. Again, $f$ is continuous on $[5,7]$ and differentiable on $(5,7)$. The MVT can be applied.

III. $[1,5]$

This interval contains $4$, where the function is undefined. So $f$ is not continuous on $[1,5]$, and the MVT cannot be applied.

Step 3: match the answer choice

Since I and II work, but III does not, the correct choice is C. I and II only.

Why this works

For rational functions, the fastest method is to inspect the denominator first. If an interval crosses a vertical asymptote or any point where the function is undefined, the MVT is immediately off the table. You do not need to compute derivatives unless the interval passes the continuity test.

A second useful habit is to read the justification carefully. The statement “because $f$ is continuous on the interval and differentiable on the open interval” is exactly the hypothesis check needed for the MVT. If either condition fails at even one point, the theorem does not apply.

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Pitfalls the pros know 👇 Students often say an interval works if the function is differentiable “almost everywhere” inside it. That is not enough for the Mean Value Theorem. The theorem requires continuity on the entire closed interval and differentiability on the entire open interval. Here, interval III fails because it crosses $x=4$, where the denominator is zero. Another common slip is forgetting that a closed interval includes its endpoints, but for a rational function the endpoints are usually not the real issue; interior singularities are the real blocker. Always check the denominator before anything else.

What if the problem changes? If the interval were changed to $[0,3]$, the MVT would still work because the interval avoids both $-2$ and $4$. If it were changed to $[-3,-1]$, it would fail because the interval contains $-2$, where the function is undefined. A useful general rule is: for $\frac{1}{(x-a)(x-b)}$, the MVT can be applied on a closed interval only if that interval stays completely away from $a$ and $b$.

Tags: continuity on a closed interval, differentiability on an open interval, vertical asymptote