If h = f-g, then it is continuous on [a,b] and differentiable on (a,b) (as you pointed out). So, by the MVT, there exists c in (a,b) s.t.
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Since f'(x)<g'(x), h'(c) is negative. Therefore, h(b) must be negative, or f(b)<g(b).
suppose that 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). [hint: Apply the Mean Value Theorem to the function h=f-g.]
I know that h=f-g is continuous on [a,b] and differentiable on (a,b), but i dont know where to go from there. If you could help me it would be greatly appreciated. Thank you!