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Why do you need continuity to apply the Mean Value Theorem? Construct a counterexample.

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Why do you need differentiability to apply the Mean Value Theorem? Find a counterexample.

One example is f ( x ) = | x | + 3 , −2 x 2

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When are Rolle’s theorem and the Mean Value Theorem equivalent?

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If you have a function with a discontinuity, is it still possible to have f ( c ) ( b a ) = f ( b ) f ( a ) ? Draw such an example or prove why not.

Yes, but the Mean Value Theorem still does not apply

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For the following exercises, determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer.

y = 1 x 3

( , 0 ) , ( 0 , )

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y = x 2 4

( , −2 ) , ( 2 , )

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For the following exercises, graph the functions on a calculator and draw the secant line that connects the endpoints. Estimate the number of points c such that f ( c ) ( b a ) = f ( b ) f ( a ) .

[T] y = 3 x 3 + 2 x + 1 over [ −1 , 1 ]

2 points

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[T] y = tan ( π 4 x ) over [ 3 2 , 3 2 ]

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[T] y = x 2 cos ( π x ) over [ −2 , 2 ]

5 points

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[T] y = x 6 3 4 x 5 9 8 x 4 + 15 16 x 3 + 3 32 x 2 + 3 16 x + 1 32 over [ −1 , 1 ]

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For the following exercises, use the Mean Value Theorem and find all points 0 < c < 2 such that f ( 2 ) f ( 0 ) = f ( c ) ( 2 0 ) .

f ( x ) = cos ( 2 π x )

c = 1 2 , 1 , 3 2

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f ( x ) = ( x 1 ) 10

c = 1

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For the following exercises, show there is no c such that f ( 1 ) f ( −1 ) = f ( c ) ( 2 ) . Explain why the Mean Value Theorem does not apply over the interval [ −1 , 1 ] .

f ( x ) = | x 1 2 |

Not differentiable

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f ( x ) = | x |

Not differentiable

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f ( x ) = x ( Hint : This is called the floor function and it is defined so that f ( x ) is the largest integer less than or equal to x . )

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For the following exercises, determine whether the Mean Value Theorem applies for the functions over the given interval [ a , b ] . Justify your answer.

y = e x over [ 0 , 1 ]

Yes

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y = ln ( 2 x + 3 ) over [ 3 2 , 0 ]

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f ( x ) = tan ( 2 π x ) over [ 0 , 2 ]

The Mean Value Theorem does not apply since the function is discontinuous at x = 1 4 , 3 4 , 5 4 , 7 4 .

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y = 9 x 2 over [ −3 , 3 ]

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y = 1 | x + 1 | over [ 0 , 3 ]

Yes

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y = x 3 + 2 x + 1 over [ 0 , 6 ]

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y = x 2 + 3 x + 2 x over [ −1 , 1 ]

The Mean Value Theorem does not apply; discontinuous at x = 0 .

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y = x sin ( π x ) + 1 over [ 0 , 1 ]

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y = ln ( x + 1 ) over [ 0 , e 1 ]

Yes

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y = x sin ( π x ) over [ 0 , 2 ]

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y = 5 + | x | over [ −1 , 1 ]

The Mean Value Theorem does not apply; not differentiable at x = 0 .

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For the following exercises, consider the roots of the equation.

Show that the equation y = x 3 + 3 x 2 + 16 has exactly one real root. What is it?

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Find the conditions for exactly one root (double root) for the equation y = x 2 + b x + c

b = ± 2 c

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Find the conditions for y = e x b to have one root. Is it possible to have more than one root?

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For the following exercises, use a calculator to graph the function over the interval [ a , b ] and graph the secant line from a to b . Use the calculator to estimate all values of c as guaranteed by the Mean Value Theorem. Then, find the exact value of c , if possible, or write the final equation and use a calculator to estimate to four digits.

[T] y = tan ( π x ) over [ 1 4 , 1 4 ]

c = ± 1 π cos −1 ( π 2 ) , c = ± 0.1533

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[T] y = 1 x + 1 over [ 0 , 3 ]

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[T] y = | x 2 + 2 x 4 | over [ −4 , 0 ]

The Mean Value Theorem does not apply.

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[T] y = x + 1 x over [ 1 2 , 4 ]

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[T] y = x + 1 + 1 x 2 over [ 3 , 8 ]

1 2 c + 1 2 c 3 = 521 2880 ; c = 3.133 , 5.867

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At 10:17 a.m., you pass a police car at 55 mph that is stopped on the freeway. You pass a second police car at 55 mph at 10:53 a.m., which is located 39 mi from the first police car. If the speed limit is 60 mph, can the police cite you for speeding?

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Two cars drive from one spotlight to the next, leaving at the same time and arriving at the same time. Is there ever a time when they are going the same speed? Prove or disprove.

Yes

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Show that y = sec 2 x and y = tan 2 x have the same derivative. What can you say about y = sec 2 x tan 2 x ?

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Show that y = csc 2 x and y = cot 2 x have the same derivative. What can you say about y = csc 2 x cot 2 x ?

It is constant.

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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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