Finding a Point of Equality in a Twice Differentiable Function on [0,1]

[kathrynag]

Let g:[0,1]-->R be twice differentiable(both g and g' are differentiable functions) with g''(x)>0 for all x in [0,1]. If g(0)>0 and g(1)=1, show that g(d)=d for some point d in (0,1) iff g'(1)>1.


I thought I might use the MVT.
g'(c)=g(1)-g(0)/1=1-g(0)
g'(c)<0 then

 

[Mark44]

It might be helpful to sketch a graph of a function that meets these criteria. The graph should be concave up, have a y intercept that is positive, and g(1) = 1.
You have to show that the graph of g crosses the line y = x iff g'(1) > 1.

 

[kathrynag]

Ok so I assume g'(1)>1
Then g'(1)=lim[(g(x)-g(1)/(1-x)}=lim[(g(x)-1)/(1-x)]>1

 

[Mark44]

Assume g'(1) > 1, then show that for some d in (0, 1), g(d) = d.

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.

 

[kathrynag]

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

 

[Mark44]

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