- #1

- 598

- 0

I thought I might use the MVT.

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

g'(c)<0 then

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter kathrynag
- Start date

- #1

- 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

Mentor

- 34,907

- 6,651

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

- #3

- 598

- 0

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

Mentor

- 34,907

- 6,651

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

- 598

- 0

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

Mentor

- 34,907

- 6,651

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

Share: