0
$\begingroup$

So let me pose the complete question:

Show the following:

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.

$\endgroup$
1
  • 3
    $\begingroup$ Apply the MVT to the difference $f-g$ .... $\endgroup$
    – Martin R
    Nov 7, 2021 at 19:11

1 Answer 1

2
$\begingroup$

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, $h(x)=h(a)+h'(\xi)(x-a)\geq0$.

$\endgroup$
3
  • 1
    $\begingroup$ Always wrong when answering a question t9 say “it is trivial.” Reads as judgmental, even if not meant as such. $\endgroup$ Nov 7, 2021 at 19:13
  • $\begingroup$ Why is $\xi$ from the interval $(a,x)$ and not $(a,b)? $\endgroup$
    – yousafe007
    Nov 7, 2021 at 19:29
  • $\begingroup$ @ yousafe007 We are applying Mean Value Theorem for $h$ on $[a,x]$. $\endgroup$ Nov 7, 2021 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .