Explanation Given: The value of f ( 2 ) is 5 and f ′ ( x ) ≥ 10 for x > 2 . Concept used: The mean value theorem: The mean value theorem states that if the function f is continuous on the interval [ a , b ] and differentiable on the interval ( a , b ) then a number c exists in the interval such that the slope at c can be expressed as follows: f ′ ( c ) = f ( b ) − f ( a ) b − a ...... (1) Proof: Mean value theorem is applied on a function which is differentiable on the interval ( a , b ) such that the value of a and b are well defined. So, for the inequality x > 2 the corresponding interval is ( 2 , x ) . Now, the mean value theorem can be applied on the interval [ 2 , x ]

4th Edition

ISBN: 9781319379421

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.3, Problem 71E

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That f(x)≥10x−15 for all x>2 if f(2)=5 and f′(x)≥10 for x>2.

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Chapter 4.3, Problem 70E

Chapter 4.3, Problem 72E

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Chapter 4 Solutions

CALCULUS 4E (HC) W/ ACHIEVE ACCESS

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That f ( x ) ≥ 10 x − 15 for all x > 2 if f ( 2 ) = 5 and f ′ ( x ) ≥ 10 for x > 2 .

Solution Summary: The author explains that the mean value theorem is applied on a function which is differentiable on the interval left[a,bright].

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That f(x)≥10x−15 for all x>2 if f(2)=5 and f′(x)≥10 for x>2.

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Chapter 4 Solutions

That f ( x ) ≥ 10 x − 15 for all x &gt; 2 if f ( 2 ) = 5 and f ′ ( x ) ≥ 10 for x &gt; 2 .

Solution Summary: The author explains that the mean value theorem is applied on a function which is differentiable on the interval left[a,bright].

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To show: That f(x)≥10x−15 for all x>2 if f(2)=5 and f′(x)≥10 for x>2.

Want to see the full answer?

Chapter 4 Solutions

That f ( x ) ≥ 10 x − 15 for all x > 2 if f ( 2 ) = 5 and f ′ ( x ) ≥ 10 for x > 2 .

To show:

That f(x)≥10x−15 for all x>2 if f(2)=5 and f′(x)≥10 for x>2.