104-so7 - Math 104 - Lecture 1 - Problem Set 7 Solutions...

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Unformatted text preview: Math 104 - Lecture 1 - Problem Set 7 Solutions 3.3a (2 points) Let x < y be given. By the mean value theorem, we knwo there is a ( x,y ) so that f ( x )- f ( y ) x- y = f ( ) > . Since x- y < , it must also be that f ( x )- f ( y ) < . That is f ( x ) < f ( y ) . 3.3b (1 point) The same argument as in (3.3a) tells us that if x < y then f ( x ) f ( y ) . So f is nondecreasing. 3.5 (2 points) Yes. Because f is continuous, we can apply LH opitals rule (continuity guarantees the numerator is zero): f (0) = lim h f ( x + h )- f ( x ) h = lim h f ( x + h ) 1 = L 3.13a (3 points) Put g ( x ) = f ( x )- x . Then f ( x ) = x if and only if g ( x ) = 0 . Also note that g ( x ) = f ( x )- 1 < L- 1 < . If g ( x ) = 0 , then we are done. If g ( x ) 6 = 0 , Put a =- g (0) L- 1 6 = 0 . By the mean value theorem, there is some between and a so that g ( a )- g (0) a = g ( ) < L- 1 =- g (0) a . So g ( a ) a < . This means g ( a ) and a have different signs. So also g (...
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104-so7 - Math 104 - Lecture 1 - Problem Set 7 Solutions...

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