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### Course: APยฎ๏ธ Calculus AB (2017 edition)ย >ย Unit 5

Lesson 2: Mean value theorem- Mean value theorem
- Conditions for MVT: graph
- Conditions for MVT: table
- Establishing differentiability for MVT
- Conditions for MVT: graph
- Justification with the mean value theorem
- Mean value theorem example: polynomial
- Mean value theorem example: square root function
- Using the mean value theorem
- Mean value theorem application
- Mean value theorem review

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# Establishing differentiability for MVT

A function must be differentiable for the mean value theorem to apply. Learn why this is so, and how to make sure the theorem can be applied in the context of a problem.

The mean value theorem (MVT) is an existence theorem similar the intermediate and extreme value theorems (IVT and EVT). Our goal is to understand the mean value theorem and know how to apply it.

## MVT and its conditions

The mean value theorem guarantees, for a function $f$ that's differentiable over an interval from $a$ to $b$ , that there exists a number $c$ on that interval such that ${f}^{\u0e42\x80\u0e12}(c)$ is equal to the function's average rate of change over the interval.

Graphically, the theorem guarantees that an arc between two endpoints has a point at which the tangent to the arc is parallel to the secant through its endpoints.

The precise conditions under which MVT applies are that $f$ is differentiable over the open interval $(a,b)$ and continuous over the closed interval $[a,b]$ . Since differentiability implies continuity, we can also describe the condition as being differentiable over $(a,b)$ and continuous at $x=a$ and $x=b$ .

Using parameters like $a$ and $b$ and talking about open and closed intervals is important if we want to be mathematically precise, but these conditions essentially mean this:

For MVT to apply, the function must be differentiable over the relevant interval, and continuous at the interval's edges.

### Why differentiability over the interval is important.

To understand why this condition is important, consider function $f$ . The function has a sharp turn between $x=a$ and $x=b$ , so it's not differentiable over $(a,b)$ .

Indeed, the function has only two possible tangent lines, neither of which is parallel to the secant between $x=a$ and $x=b$ .

### Why continuity at the edges is important.

To understand this, consider function $g$ .

As long as $g$ is differentiable over $(a,b)$ and continuous at $x=a$ and $x=b$ , MVT applies.

Now let's change $g$ so it's not continuous at $x=b$ . In other words, the one-sided limit $\underset{x\u0e42\x86\x92{b}^{\u0e42\x88\x92}}{lim}g(x)$ remains the same, but the function value changes to something else.

Notice how all of the possible tangent lines on the interval are necessarily increasing, while the secant line is decreasing. So there isn't any tangent line that's parallel to the secant line.

In general, if a function isn't continuous at the edges, the secant line will be disconnected from the tangent lines along the interval.

In Problem set 1 we will analyze the applicability of the mean value theorem to function $h$ at different intervals.

*Want more practice? Try this exercise.*

**Notice:**When MVT doesn't apply, all we can tell is that we aren't certain the conclusion is true. It does

*not*mean that the conclusion

*isn't*true.

In other words, it's possible to have a point where the tangent is parallel to the secant, even when MVT doesn't apply. We just can't be

*certain*about it unless the conditions for MVT have been met.For example, in the last problem, MVT didn't apply to $f$ over the interval $[\u0e42\x88\x921,5]$ , even though there are actually two points in the interval $[\u0e42\x88\x921,5]$ where the tangent is parallel to the secant between the endpoints.

*Want more practice? Try this exercise.*

### Common mistake: Not recognizing when the conditions are met

Let's take Problem 3 for example. These are the common ways we would expect the conditions for MVT to look:

is differentiable over$h$ and continuous over$(3,7)$ .$[3,7]$ is differentiable over$h$ and continuous at$(3,7)$ and$x=3$ .$x=7$

However, we will not always be given information about the function this way. For example, if $h$ is differentiable over $[3,7]$ , the conditions are met because differentiability implies continuity.

Another example is when $h$ is differentiable over a larger interval, for example $(2,8)$ . Even though continuity isn't mentioned, differentiability over $(2,8)$ implies differentiability over $(3,7)$ and continuity over $[3,7]$ .

### Common mistake: Applying the wrong existence theorem

By now, we are familiar with three different existence theorems: the intermediate value theorem (IVT), the extreme value theorem (EVT), and the mean value theorem (MVT). They have a similar structure but they apply under different conditions and guarantee different kinds of points.

- IVT guarantees a point where the
*function*has a certain value between two given values. - EVT guarantees a point where the
*function*obtains a maximum or a minimum value. - MVT guarantees a point where the
*derivative*has a certain value.

Before applying one of the existence theorems, make sure you understand the problem well enough to tell which theorem should be applied.

## Want to join the conversation?

- Respected Sir,

I have a doubt under "Why continuity at the edges is important.".

I did not quite understand in the assumption that is considered, on why is it, that the secant line doesn't join to a point in the curve considered? Isn't it supposed to join to some point before the point b in the curve?(11 votes)- you're thinking about a tangent, a secant is simply a line, from point a to point b(15 votes)

- Things seem a bit out of order ... This article refers to the EVT, but the EVT is not introduced until the next lesson.(17 votes)
- What is the Extreme Value Theorem?(6 votes)
- It's the fact that a continuous function on a closed interval will achieve a maximum value and a minimum value.(10 votes)

- Why is the below worded as it is. Specifically why not just say that f(x) is differentiable over [a, b]? How is the statement I said different different then the statement below? Why does it matter?

My theory is that the end of a line can't be differntiable but i don't know.

"The precise conditions under which MVT applies are that fff is differentiable over the open interval (a,b) and continuous over the closed interval [a,b]. Since differentiability implies continuity, we can also describe the condition as being differentiable over (a,b) and continuous at x=a and x=b."(4 votes)- Your suspicion is correct. We can't differentiate the function at the end points of a closed interval.

See rashaveraka's answer to this question:

https://www.khanacademy.org/math/ap-calculus-ab/ab-existence-theorems/ab-mvt/v/mean-value-theorem-1?qa_expand_key=kaencrypted_2edb39dd806f4a066e87def729042ba5_bdc4f8623227f4d08c2ef1e0f0434660194306685ca9cbbb58337efd9174b0146234d247078dc05fb582dcb55d4dd5843fe41cb067f9826a1ffeea2112cf2c88239f510de888397343745ca05bff49bdefed5141cc558eaf3c818bd9f02ca567c70bcc3fc621e988078de6256e9333e17ac868f66fcdf5835e90eec38fc7bf187b9bef4b0bb8d1552bfd891fa34a90a3d2a7288854c26b4532ac73fdf4c4ac3d(5 votes)

- In problem 3, the question "Which condition makes James's claim true?" is a little misleading. It assumes there is something in the information given that makes the claim true, but there isn't enough information to make that determination. A better question would be "Which condition WOULD MAKE James's claim true?"(5 votes)
- Wait, I don't remember learning about IVT and EVT yet. This page said, "By now, we are familiar with three different existence theorems: the intermediate value theorem (IVT), the extreme value theorem (EVT), and the mean value theorem (MVT)."

Does anyone know which unit that is in?(5 votes)- IVT was in "Limits and Continuity" and EVT was in "Applying Derivatives to analyze functions."(0 votes)

- Great article. I understand why sharp turns would ruin the MVT, but how vertical tangent does?(1 vote)
- Remember what the Mean Value Theorem's conditions are. It says that a function needs to be
**differentiable**on some interval (a,b). Now, if we have a vertical tangent in said interval, as the slope of the tangent is undefined, our interval is no longer differentiable, as we have an undefined slope at a point. Hence, vertical tangents would make MVT invalid on the interval(2 votes)

- In the first MVT video Sal says that the rules for MVT are that it is continuous for the entire interval [a, b] but only has to be differentiable for the interval inside (a,b).

Then, in the first section of questions on this article, it says "hhh is differentiable over the entire interval between x=-3x=โ3x, equals, minus, 3 and x=2x=2x, equals, 2, but it isn't continuous at x=-3x=โ3x, equals, minus, 3.

The answer is no."

Which one is right?(1 vote) - Why does the Mean Value Theorem specify differentiability on an open interval (a,b) rather than a half-open interval like [a,b] or (a,b]? What implications would including an endpoint have on the theorem's validity?(1 vote)
- We don't consider open intervals because endpoints aren't differentiable. Differentiability requires the derivative to be same on both sides of a point. But for endpoints, we don't have "both sides"(1 vote)

- Hi,

If I do not draw the graph of a function that has a sharp turn, will there be another way to determine whether or not there is a sharp turn or do I have to draw the graph?(1 vote)- Well, you could always check for differentiability. Differentiate the function, and see where it isn't defined. Sure, the point where it isn't defined needn't necessarily be a sharp turn, but that at least proves that there's a non-differentiable point in the function.(1 vote)