\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Derivatives - Mean Value Theorem

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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, \(x\).
2. 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.
3. 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.

What This Theorem Gives Us

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

Plot 1

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.

Graphically, it looks like what we show here in Plot 1.
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 [min-secs]

video by MIT OCW

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 [14min-36secs]

video by PatrickJMT

beijing

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

Okay, time for some practice problems.

Practice

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

Basic

\(f(x)=3x^2+6x-5\);   \([-2,1]\)

Problem Statement

\(f(x)=3x^2+6x-5\);   \([-2,1]\)

Solution

46 video

video by Krista King Math

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\(f(x)=x^2-2x\);   \([0,2]\)

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=x^2-2x\) and interval \([0,2]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1147 video

video by Krista King Math

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\(f(x)=\sqrt{x}-x/3\);   \([0,9]\)

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function and \(f(x)=\sqrt{x}-x/3\) interval \([0,9]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1150 video

video by Krista King Math

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\(f(x)=x^{2/3}\);   \([0,1]\)

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=x^{2/3}\) and interval \([0,1]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1151 video

video by MIP4U

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\(f(x)=x^2-3x+2\);   \([1,2]\)

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=x^2-3x+2\) and interval \([1,2]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1153 video

video by MIP4U

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\(f(x)=(x^2-2x)e^x\);   \([0,2]\)

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=(x^2-2x)e^x\) and interval \([0,2]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1154 video

video by MIP4U

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

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=\sin(2x)\) and interval \([\pi/6,\pi/3]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1155 video

video by MIP4U

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Intermediate

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

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=2\sin(x)\cos(x)\) and interval \([0,\pi]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1148 video

video by Krista King Math

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\(f(x)=1-|x|\);   \([-1,1]\)

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(f(x)=1-|x|\) and interval \([-1,1]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1149 video

video by Krista King Math

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

Problem Statement

Determine all numbers \(c\) which satisfy the Mean Value Theorem (or Rolles Theorem) for the function \(\displaystyle{f(x)=\frac{x}{x+1}}\) and interval \([-1/2,2]\), if possible. If the theorem does not apply, explain why. Give your answer in exact form.

Solution

1152 video

video by MIP4U

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

mean value theorem youtube playlist

WikiBooks: Mean Value Theorem (less detail)

Wikipedia: Mean Value Theorem (more detail with advanced discussion)

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Mean Value Theorem

What This Theorem Requires

What This Theorem Gives Us

Practice

Calculus: Early Transcendental Functions

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Practice Instructions

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

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