104-so7

# 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 &lt; y be given. By the mean value theorem, we knwo there is a ( x,y ) so that f ( x )- f ( y ) x- y = f ( ) &gt; . Since x- y &lt; , it must also be that f ( x )- f ( y ) &lt; . That is f ( x ) &lt; f ( y ) . 3.3b (1 point) The same argument as in (3.3a) tells us that if x &lt; 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 &lt; L- 1 &lt; . 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 ( ) &lt; L- 1 =- g (0) a . So g ( a ) a &lt; . 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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