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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(c) is equal to the function's average rate of change over the interval.
f(c)=f(b)f(a)ba
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.
A function is graphed. The positive x-axis is unmarked. The graph is a curve. The curve starts at a closed circle at the origin, moves upward to a peak, moves downward slightly, and ends at a closed circle in quadrant 1. A secant line connects the end points of the curve. A tangent line is drawn parallel to the secant line, and touches the curve somewhere between the end points.
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).
Function f is graphed. The positive x-axis includes values a and b, from left to right. The graph is a set of line segments. The set starts at a closed circle above x = a, moves upward to a sharp turn, moves downward, and ends at a closed circle at x = b, higher than the starting point.
Indeed, the function has only two possible tangent lines, neither of which is parallel to the secant between x=a and x=b.
The graph of function f has 2 tangent lines and a secant line. Tangent A starts in quadrant 3, moves upward along the upward line segment, and ends in quadrant 1. Tangent B starts in quadrant 1, moves downward along the downward line segment, and ends in quadrant 1. A secant line connects the endpoints of the line segments.

Why continuity at the edges is important.

To understand this, consider function g.
Function g is graphed. The positive x-axis includes values a and b, from left to right. The graph is a curve. The curve starts at a closed circle above x = a, moves upward with increasing steepness, and ends at a closed circle at x = b.
As long as g is differentiable over (a,b) and continuous at x=a and x=b, MVT applies.
The graph of function g has a tangent line and a secant line. The tangent line starts in quadrant 4, moves upward, touches the curve, and ends in quadrant 1. The secant line connects the endpoints of the curve.
Now let's change g so it's not continuous at x=b. In other words, the one-sided limit limxbg(x) remains the same, but the function value changes to something else.
Function g is graphed. The positive x-axis includes values a and b, from left to right. The graph is a curve. The curve starts at a closed circle above x = a, moves upward with increasing steepness, and ends at an open circle at x = b. A closed point is plotted at x = b, below the starting point at x = a. A secant line connects the 2 closed circles.
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.
Problem 1.A
Function h is graphed. The x-axis goes from negative 8 to 8. The graph consists of a curve and a set of line segments. The curve starts in quadrant 1, moves downward, moves vertically through (negative 6, 3), and ends at a closed circle at (negative 3, negative 3). The set starts at an open circle at (negative 3, negative 5), moves upward to a sharp turn at (6, 4), moves downward, and ends at (8, 2).
Does MVT apply to h over the interval [5,1]?
Choose 1 answer:

Problem 2
The graph of function f has a vertical tangent at x=2.
Does MVT apply to f over the interval [1,5]?
Choose 1 answer:

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 [1,5], even though there are actually two points in the interval [1,5] where the tangent is parallel to the secant between the endpoints.
Function f is graphed. The x-axis goes from negative 2 to 8. The graph consists of a curve. The curve starts in quadrant 3, moves downward to point (negative 1, negative 3), moves upward, moving vertically through (2, 0), continues to point (5, 3), moves downward, and ends at (8, 0). Two parallel tangent lines each starts in quadrant 3, moves upward, and ends in quadrant 1. The upper tangent line touches the curve at about (2.8, 2.2). The lower tangent line touches the curve at about (1.2, negative 2.2). A secant line connects points (negative 1, negative 3) and (5, 3).
Problem 3
This table gives a few values of function h.
x371011
h(x)1526
James said that since h(7)h(3)73=1 there must be a number c in the interval [3,7] for which h(c)=1.
Which condition makes James's claim true?
Choose 1 answer:

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:
  • h is differentiable over (3,7) and continuous over [3,7].
  • h is differentiable over (3,7) and continuous at x=3 and 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].
Problem 4
f is a differentiable function. f(1)=2 and f(5)=2.
Match each conclusion with its appropriate existence theorem.
1

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?

  • blobby green style avatar for user Anupama
    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)
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  • blobby green style avatar for user Chris Trimble
    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)
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  • leaf green style avatar for user Seth
    What is the Extreme Value Theorem?
    (6 votes)
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  • piceratops ultimate style avatar for user Garret Cervantez
    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)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user JSperling
    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)
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  • blobby green style avatar for user 2VETURIrithvik
    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)
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  • male robot johnny style avatar for user Mohamed Ibrahim
    Great article. I understand why sharp turns would ruin the MVT, but how vertical tangent does?
    (1 vote)
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    • male robot donald style avatar for user Venkata
      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)
  • piceratops tree style avatar for user Connie Williamson
    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=−3x, equals, minus, 3 and x=2x=2x, equals, 2, but it isn't continuous at x=-3x=−3x, equals, minus, 3.
    The answer is no."

    Which one is right?
    (1 vote)
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  • blobby green style avatar for user aphinoo.damian
    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)
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  • blobby green style avatar for user 177782
    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)
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    • male robot donald style avatar for user Venkata
      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)