Can someone explain the steps to me please

[darius]

the question is if f and g are continuous on [a,b] and differentiable on (a,b). Suppose also that f(a)=g(a) and f'(x)<g'(x) for a<x<b. Prove that
f(b)<g(b). The hint is to apply the mean value theorem to the function
h=f-g
So assume that h=f-g. Since f and g are continuous on [a,b] and differentiable on (a,b) the mean value theorem applies by which there has to be a number such that h(b)= h(b)-h(a)=h'(c)(b-a). Since h'(c)<0 then h'(c)(b-a)<0, so
f(b)-g(b)=h(b)<0 and hence f(b)<g(b).
in this explanation why is h(b)=h(b)-h(a) and why is h'(c)<O.


The folly of mistaking a paradox for a discovery, a metaphor for a proof, a torrent of verbiage for a torrent of capital truths, and ourselves for an oracle, is inborn in us--Paul Valerey

 

[StatusX]

h(a)=f(a)-g(a)=0, and h'(c)<0 since h'(x)=f'(x)-g'(x)<0 for all x between a and b.

 

[darius]

Please bear with me. I am new to calculus. Will be starting college in the fall and am learning now on my own. Thank you for your help status!