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