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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, prime, left parenthesis, c, right parenthesis is equal to the function's average rate of change over the interval.
f, prime, left parenthesis, c, right parenthesis, equals, start fraction, f, left parenthesis, b, right parenthesis, minus, f, left parenthesis, a, right parenthesis, divided by, b, minus, a, end fraction
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 left parenthesis, a, comma, b, right parenthesis and continuous over the closed interval open bracket, a, comma, b, close bracket. Since differentiability implies continuity, we can also describe the condition as being differentiable over left parenthesis, a, comma, b, right parenthesis and continuous at x, equals, a and x, equals, 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, equals, a and x, equals, b, so it's not differentiable over left parenthesis, a, comma, b, right parenthesis.
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, equals, a and x, equals, 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 left parenthesis, a, comma, b, right parenthesis and continuous at x, equals, a and x, equals, 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, equals, b. In other words, the one-sided limit limit, start subscript, x, \to, b, start superscript, minus, end superscript, end subscript, g, left parenthesis, x, right parenthesis 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
• Current
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 open bracket, minus, 5, comma, minus, 1, close bracket?

Problem 2
The graph of function f has a vertical tangent at x, equals, 2.
Does MVT apply to f over the interval open bracket, minus, 1, comma, 5, close bracket?

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 open bracket, minus, 1, comma, 5, close bracket, even though there are actually two points in the interval open bracket, minus, 1, comma, 5, close bracket 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, left parenthesis, x, right parenthesis15minus, 2minus, 6
James said that since start fraction, h, left parenthesis, 7, right parenthesis, minus, h, left parenthesis, 3, right parenthesis, divided by, 7, minus, 3, end fraction, equals, 1 there must be a number c in the interval open bracket, 3, comma, 7, close bracket for which h, prime, left parenthesis, c, right parenthesis, equals, 1.
Which condition makes James's claim true?

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 left parenthesis, 3, comma, 7, right parenthesis and continuous over open bracket, 3, comma, 7, close bracket.
• h is differentiable over left parenthesis, 3, comma, 7, right parenthesis and continuous at x, equals, 3 and x, equals, 7.
However, we will not always be given information about the function this way. For example, if h is differentiable over open bracket, 3, comma, 7, close bracket, the conditions are met because differentiability implies continuity.
Another example is when h is differentiable over a larger interval, for example left parenthesis, 2, comma, 8, right parenthesis. Even though continuity isn't mentioned, differentiability over left parenthesis, 2, comma, 8, right parenthesis implies differentiability over left parenthesis, 3, comma, 7, right parenthesis and continuity over open bracket, 3, comma, 7, close bracket.
Problem 4
f is a differentiable function. f, left parenthesis, 1, right parenthesis, equals, minus, 2 and f, left parenthesis, 5, right parenthesis, equals, 2.
Match each conclusion with its appropriate existence theorem.

### 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.