Explanation Given: The value of f ( 0 ) is 2 and f ′ ( x ) ≤ 3 . 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: The mean value theorem can be applied on the interval [ 0 , 4 ] . Substitute 4 for b , 0 for a and x for c in equation (1) as follows: f ′ ( x ) = f ( 4 ) − f ( 0 ) 4 − 0 f ′ ( x ) = f ( 4 ) − f ( 0 ) 4 It is given that f ′ ( x ) ≤ 3 then the above equation can be written as follows: f ( 4 ) − f ( 0 ) 4 ≤ 3 Substitute 2 for f ( 0 ) in the above equation as follows: f ( 4 ) − 2 4 ≤ 3 f ( 4 ) − 2 ≤ 3 ⋅ 4 f ( 4 ) − 2 ≤ 12 f ( 4 ) ≤ 14 For x > 0 , the mean value theorem is applied on the interval [ 0 , x ]

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

4th Edition

ISBN: 9781319057664

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 4.3, Problem 69E

To determine

To show:

That $f(4)≤14$ for $[0,4]$ if $f(0)=2$ and $f′(x)≤3$ for $x>0$ and also prove that $f(x)≤2+3x$ for all $x>0$ .

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Chapter 4.3, Problem 68E

Chapter 4.3, Problem 70E

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Chapter 4 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.1 - 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That f ( 4 ) ≤ 14 for [ 0 , 4 ] if f ( 0 ) = 2 and f ′ ( x ) ≤ 3 for x > 0 and also prove that f ( x ) ≤ 2 + 3 x for all x > 0 .

Solution Summary: The author explains the mean value theorem, which states that if the function f is continuous and differentiable on the interval, then the slope can be expressed as follows.

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That $f(4)≤14$ for $[0,4]$ if $f(0)=2$ and $f′(x)≤3$ for $x>0$ and also prove that $f(x)≤2+3x$ for all $x>0$ .

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Chapter 4 Solutions

That f ( 4 ) ≤ 14 for [ 0 , 4 ] if f ( 0 ) = 2 and f ′ ( x ) ≤ 3 for x &gt; 0 and also prove that f ( x ) ≤ 2 + 3 x for all x &gt; 0 .

Solution Summary: The author explains the mean value theorem, which states that if the function f is continuous and differentiable on the interval, then the slope can be expressed as follows.

Concept explainers

Videos

Want to see the full answer?

Chapter 4 Solutions

That f ( 4 ) ≤ 14 for [ 0 , 4 ] if f ( 0 ) = 2 and f ′ ( x ) ≤ 3 for x > 0 and also prove that f ( x ) ≤ 2 + 3 x for all x > 0 .