Mean Value Theorem - Revision history

Lila: /* Resources */

2021-10-28T16:02:52Z

Resources

Khanh at 17:22, 10 October 2021

2021-10-10T17:22:06Z

Khanh at 17:21, 10 October 2021

2021-10-10T17:21:20Z

Khanh at 17:20, 10 October 2021

2021-10-10T17:20:38Z

James.kercheville: Added links for PowerPoint and worksheet

2020-08-17T13:22:25Z

Added links for PowerPoint and worksheet

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* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremPwPt.pptx The Mean Value Theorem] PowerPoint file created by Dr. Sara Shirinkam, UTSA.

* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremWS1.pdf The Mean Value Theorem Worksheet]

← Older revision		Revision as of 16:02, 28 October 2021
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	* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremWS1.pdf The Mean Value Theorem Worksheet]		* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremWS1.pdf The Mean Value Theorem Worksheet]
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikibooks.org/wiki/Calculus/Mean_Value_Theorem Mean Value Theorem, Wikibooks: Calculus] under a CC BY-SA license

@@ Line 1: / Line 1: @@
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">
-:Mean Value Theorem
+:'''Mean Value Theorem'''
 :If <math>f(x)</math> is continuous on the closed interval <math>[a, b]</math> and differentiable on the open interval <math>(a,b)</math> , there exists a number <math>c\in(a,b)</math> such that
 :<math>f'(c)=\frac{f(b)-f(a)}{b-a}</math>
@@ Line 77: / Line 77: @@
 ==Cauchy's Mean Value Theorem==
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">
-:Cauchy's Mean Value Theorem
+:'''Cauchy's Mean Value Theorem'''
 :If <math>f(x),g(x)</math> are continuous on the closed interval <math>[a,b]</math> and differentiable on the open interval <math>(a,b)</math> , <math>g(a)\ne g(b)</math> and <math>g'(x)\ne0</math> , then there exists a number <math>c\in(a,b)</math> such that
 :<math>\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}</math>

@@ Line 4: / Line 4: @@
 :If <math>f(x)</math> is continuous on the closed interval <math>[a, b]</math> and differentiable on the open interval <math>(a,b)</math> , there exists a number <math>c\in(a,b)</math> such that
 :<math>f'(c)=\frac{f(b)-f(a)}{b-a}</math>
+</blockquote>
 ==Examples==
@@ Line 80: / Line 80: @@
 :If <math>f(x),g(x)</math> are continuous on the closed interval <math>[a,b]</math> and differentiable on the open interval <math>(a,b)</math> , <math>g(a)\ne g(b)</math> and <math>g'(x)\ne0</math> , then there exists a number <math>c\in(a,b)</math> such that
 :<math>\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}</math>
 ==Resources==

← Older revision		Revision as of 17:20, 10 October 2021
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		+
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	:Mean Value Theorem
		+	:If <math>f(x)</math> is continuous on the closed interval <math>[a, b]</math> and differentiable on the open interval <math>(a,b)</math> , there exists a number <math>c\in(a,b)</math> such that
		+	:<math>f'(c)=\frac{f(b)-f(a)}{b-a}</math>
		+
		+
		+	==Examples==
		+
		+	[[Image:Mvt2.svg\|right\|thumb\|250px]]
		+	What does this mean? As usual, let us utilize an example to grasp the concept. Visualize (or graph) the function <math>f(x)=x^3</math> . Choose an interval (anything will work), but for the sake of simplicity, [0,2]. Draw a line going from point (0,0) to (2,8). Between the points <math>x=0</math> and <math>x=2</math> exists a number <math>x=c</math> , where the derivative of <math>f</math> at point <math>c</math> is equal to the slope of the line you drew.
		+
		+	;Solution
		+	1: Using the definition of the mean value theorem
		+
		+	:<math>\frac{f(b)-f(a)}{b-a}</math>
		+
		+	insert values. Our chosen interval is [0,2]. So, we have
		+
		+	:<math>\frac{f(2)-f(0)}{2-0}=\frac82=4</math>
		+
		+	2: By the definition of the mean value theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point <math>x=c</math> .
		+
		+	:<math>\frac{dy}{dx}=3x^2</math>
		+
		+	Now, we know that the slope of the point is 4. So, the derivative at this point <math>c</math> is 4. Thus, <math>4=3x^2</math> . The square root of 4/3 is the point.
		+
		+
		+	'''Example 2:''' Find the point that satisifes the mean value theorem on the function <math>f(x)=\sin(x)</math> and the interval <math>[0,\pi]</math> .
		+	;Solution
		+	1: Always start with the definition:
		+
		+	:<math>\frac{f(b)-f(a)}{b-a}</math>
		+
		+	so,
		+
		+	:<math>\frac{\sin(\pi)-\sin(0)}{\pi-0}=0</math>
		+
		+	(Remember, <math>\sin(\pi)</math> and <math>\sin(0)</math> are both 0.)
		+
		+	2: Now that we have the slope of the line, we must find the point <math>x=c</math> that has the same slope. We must now get the derivative!
		+
		+	:<math>\frac{d\sin(x)}{dx}=\cos(x)=0</math>
		+
		+	The cosine function is 0 at <math>\frac{\pi}{2}+k\pi </math> , where <math>k</math> is an integer. Remember, we are bound by the interval <math>[0,\pi]</math> , so <math>\frac{\pi}{2}</math> is the point <math>c</math> that satisfies the Mean Value Theorem.
		+
		+	==Differentials==
		+	<!-- I don't think my explanation is very lucid...would someone like to explain it better? Also, I don't know how to synthesize capital Deltas. Thanks! -EbolaPox !-->
		+	Assume a function <math>y=f(x)</math> that is differentiable in the open interval <math>(a,b)</math> that contains <math>x</math> . <math>\Delta y=\frac{dy}{dx}\cdot\Delta x</math>
		+
		+	The "Differential of <math>x</math>" is the <math>\Delta x</math> . This is an approximate change in <math>x</math> and can be considered "equivalent" to <math>dx</math> . The same holds true for <math>y</math> . What is this saying? One can approximate a change in <math>y</math> by knowing a change in <math>x</math> and a change in <math>x</math> at a point very nearby. Let us view an example.
		+
		+	Example: A schoolteacher has asked her students to discover what <math>4.1^2</math> is. The students, bereft of their calculators, are too lazy to multiply this out by hand or in their head and desire to utilize calculus. How can they approximate this?
		+
		+	1: Set up a function that mimics the procedure. What are they doing? They are taking a number (Call it <math>x</math>) and they are squaring it to get a new number (call it <math>y</math>). Thus, <math>y=x^2</math> Write yourself a small chart. Make notes of values for <math>x,y,\Delta x,\Delta y,\frac{dy}{dx}</math> . We are seeking what <math>y</math> really is, but we need the change in <math>y</math> first.
		+
		+	2: Choose a number close by that is easy to work with. Four is very close to 4.1, so write that down as <math>x</math> . Your <math>\delta x</math> is .1 (This is the "change" in <math>x</math> from the approximation point to the point you chose.)
		+
		+	3: Take the derivative of your function.
		+
		+	<math>\frac{dy}{dx}=2x</math> . Now, "split" this up (This is not really what is happening, but to keep things simple, assume you are "multiplying" <math>dx</math> over.)
		+
		+	3b. Now you have <math>dy=2x\cdot dx</math> . We are assuming <math>dy,dx</math> are approximately the same as the change in <math>x</math> , thus we can use <math>\Delta x</math> and <math>y</math> .
		+
		+	3c. Insert values: <math>dy=2\cdot4\cdot0.1</math> . Thus, <math>dy=0.8</math> .
		+
		+	4: To find <math>F(4.1)</math> , take <math>F(4)+dy</math> to get an approximation. 16 + 0.8 = 16.8; This approximation is nearly exact (The real answer is 16.81. This is only one hundredth off!)
		+
		+	==Definition of Derivative==
		+	The exact value of the derivative at a point is the rate of change over an infinitely small distance, approaching 0. Therefore, if h approaches 0 and the function is <math>f(x)</math> :
		+
		+	:<math>f'(x)=\frac{f(x+h)-f(x)}{(x+h)-x}</math>
		+
		+	If h approaches 0, then:
		+	:<math>f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}</math>
		+
		+	==Cauchy's Mean Value Theorem==
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	:Cauchy's Mean Value Theorem
		+	:If <math>f(x),g(x)</math> are continuous on the closed interval <math>[a,b]</math> and differentiable on the open interval <math>(a,b)</math> , <math>g(a)\ne g(b)</math> and <math>g'(x)\ne0</math> , then there exists a number <math>c\in(a,b)</math> such that
		+	:<math>\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}</math>
		+
		+	==Resources==
	* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremPwPt.pptx The Mean Value Theorem] PowerPoint file created by Dr. Sara Shirinkam, UTSA.		* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremPwPt.pptx The Mean Value Theorem] PowerPoint file created by Dr. Sara Shirinkam, UTSA.

	* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremWS1.pdf The Mean Value Theorem Worksheet]		* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Mean%20Value%20Theorem/MAT1214-4.4TheMeanValueTheoremWS1.pdf The Mean Value Theorem Worksheet]