- #1

kathrynag

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I thought I might use the MVT.

g'(c)=g(1)-g(0)/1=1-g(0)

g'(c)<0 then

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- Thread starter kathrynag
- Start date

- #1

kathrynag

- 598

- 0

I thought I might use the MVT.

g'(c)=g(1)-g(0)/1=1-g(0)

g'(c)<0 then

- #2

Mark44

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You have to show that the graph of g crosses the line y = x iff g'(1) > 1.

- #3

kathrynag

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Ok so I assume g'(1)>1

Then g'(1)=lim[(g(x)-g(1)/(1-x)}=lim[(g(x)-1)/(1-x)]>1

Then g'(1)=lim[(g(x)-g(1)/(1-x)}=lim[(g(x)-1)/(1-x)]>1

- #4

Mark44

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Then go the other way - g(d) = d ==> g'(1) = 1.

My suggestion of graphing was to help you get a geometric feel for this problem.

- #5

kathrynag

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I guess I get stuck just assuming g'(1)>1

Then g'(1)=lim[(g(x)-g(1)/(1-x)}=lim[(g(x)-1)/(1-x)]>1

Then g'(1)=lim[(g(x)-g(1)/(1-x)}=lim[(g(x)-1)/(1-x)]>1

- #6

Mark44

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Let f(x) = g(x) - x. Can you do something with that?

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