# Proving $f(x) \ge g(x)$ for all $x \in [a,b]$

So let me pose the complete question:

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Let $$f, g : [a,b] \to \mathbb{R}$$ continuous and differentiable on the interval (a,b) with $$f(a) \ge g(a)$$ and $$f'(x) \ge g'(x)$$ for all $$x \in (a,b)$$, then it follows that $$f(x) \ge g(x)$$ for all $$x \in [a,b]$$

We started this semester with differentiation and MVT, but I am clueless how to approach this problem.

Now looking at the functions, it is evident that it should be true, since the $$f'(x)$$ is always greater than or equal to $$g'(x)$$, which implies the monotony of the functions and hence all values being greater in the original function. But that is as far as I could get.

• Apply the MVT to the difference $f-g$ .... Nov 7, 2021 at 19:11

It is trivial. Let $$h=f-g$$, then $$h$$ is continuous on $$[a,b]$$, differentiable on $$(a,b)$$. Moreover, $$h(a)\geq0$$ and $$h'(x)\geq0$$ for all $$x\in(a,b)$$. Let $$x\in(a,b]$$ be arbitrary. By Mean Value Theorem, there exists $$\xi\in(a,x)$$ such that $$h(x)-h(a)=h'(\xi)(x-a)$$. Therefore, $$h(x)=h(a)+h'(\xi)(x-a)\geq0$$.
• Why is $\xi$ from the interval $(a,x)$ and not $(a,b)? Nov 7, 2021 at 19:29 • @ yousafe007 We are applying Mean Value Theorem for$h$on$[a,x]\$. Nov 7, 2021 at 21:04