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17calculus > derivatives > mean value theorem

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Mean Value Theorem

The Mean Value Theorem is one of the coolest applications of the derivative. Using it doesn't introduce any new notation or concept. It is just another way to apply the derivative on a continuous function. Here is a formal version of the theorem.

Mean Value Theorem

For any function that is continuous on \([a, b]\) and differentiable on \((a, b)\) there exists some \(c\) in the interval \((a, b)\) such that the secant joining the endpoints of the interval \([a, b]\) is parallel to the tangent at \(c\).

Rolle's Theorem - Rolle's Theorem is a special case of the Mean Value Theorem where \(f(a)=f(b)\).

Let's break this down.
What This Theorem Requires
1. First, we are given a closed interval \([a,b]\). Notice that all these intervals and values of \(c\) refer to the independent variable, \(x\).
1. Second, we must have a function that is continuous on the given interval \([a,b]\). We don't care what's going on outside this interval.
2. Next, the function must be differentiable inside the open interval \((a,b)\). This means there must not be any sharp corners inside the interval. Also, we don't need to have the function differentiable at the end points, only inside the interval.

Source: Wikipedia - Mean Value Theorem

Notice that the orange and green 'lines' are drawn as vectors since they have an arrow at one end. In this context, we would usually want to draw lines and not vectors. But check with your instructor to see what they expect.

What This Theorem Gives Us
If we have met the conditions above, we are guaranteed the following.
3. We can draw a line (called the secant line) between the end points.
4. We are guaranteed that there is a tangent line to the curve that is parallel to the secant line.
5. This tangent line occurs at some \(x=c\) where \( a < c < b \).

Graphically, it looks like this.
The idea is that we are given the function (which we must check to see if it is meets conditions 1 and 2 above). Then we draw the orange secant line. We are then guaranteed that the green tangent line exists. Let's see how we can actually find this point \(x=c\) and the equation of the tangent line.

This equation should have some familiar pieces.
\( \displaystyle{ f'(c) = \frac{f(b)-f(a)}{b-a} }\)
The right side is the slope of a line through the points \((a, f(a))\) and \((b, f(b))\). So this gives us the slope of the secant line. The left side is the slope of the tangent line through the point \((c,f(c))\). So you set those equal to each other and solve for \(c\). Once you have \(c\), you can find the equation of the tangent line.

Okay, so there are a lot of equations so far but what does this mean intuitively? This first video clip will help you really understand the mean value theorem, what it is saying and where it comes from (with a proof).

MIT OCW - Lec 14 | MIT 18.01 Single Variable Calculus, Fall 2007

Here is a good video explaining the mean value theorem in some detail with a great example. The example is quite extensive but he takes the time to explain some of the things you may run across when using this theorem.

PatrickJMT - The Mean Value Theorem

Okay, time for some practice problems.

beijing

China celebrates the Mean Value Theorem. Pretty cool, eh? [ Source: Wikipedia - Mean Value Theorem ]

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Practice Problems

Instructions - - Unless otherwise instructed, determine all numbers c which satisfy the Mean Value Theorem (or Rolles Theorem) for the given functions and intervals, if possible. If the theorem does not apply, explain why. Give your answers in exact form.

Level A - Basic

Practice A01

\(f(x)=3x^2+6x-5\); \([-2,1]\)

solution

Practice A02

\(f(x)=x^2-2x\); \([0,2]\)

solution

Practice A03

\(f(x)=\sqrt{x}-x/3\); \([0,9]\)

solution

Practice A04

\(f(x)=x^{2/3}\); \([0,1]\)

solution

Practice A05

\(f(x)=x^2-3x+2\); \([1,2]\)

solution

Practice A06

\(f(x)=(x^2-2x)e^x\); \([0,2]\)

solution

Practice A07

\(f(x)=\sin(2x)\); \([\pi/6,\pi/3]\)

solution


Level B - Intermediate

Practice B01

\(f(x)=2\sin(x)\cos(x)\); \([0,\pi]\)

solution

Practice B02

\(f(x)=1-\abs{x}\); \([-1,1]\)

solution

Practice B03

\(\displaystyle{f(x)=\frac{x}{x+1}}\); \([-1/2,2]\)

solution

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