prove one function is less than other with mean value theorem

[asifrahman1988]

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.]​

 

[pka]

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.]

Apply the mean value theorem to the function \(\displaystyle h\).
\(\displaystyle (\exists c\in(a,b))\left[h'(c)=\dfrac{h(b)-h(a)}{b-a}\right]~.\)

From the given we can see that \(\displaystyle \dfrac{h(b)-h(a)}{b-a}=\dfrac{f(b)-g(b)}{b-a}\) and \(\displaystyle h'(c)=f'(c)-g'(c)<0\).

Can you finish?