Suppose f and g are continuous on [a, b] and that f(a) < g(a), but f(b) > g(b). Prove that f(x) = g(x) for some x in [a, b].

Question

I have been asked to prove the following question for my Real Analysis course, and I'm not quite sure how to go about proving this. The textbook is Spivak's Calculus, and this is question 10 from Chapter 7. Thank you in advance!

Suppose $f,g$ are continuous on $[a, b]$ and that $f(a) < g(a)$, but $f(b) > g(b)$. Prove that $f(x) = g(x)$ for some $x$ in $[a, b]$.

My idea is that this problem will involve the intermediate value theorem, as $f$ holds values both less than and greater than $g$ for some $x$.

This seems to be a duplicate of: Proving that $f(x) = g(x)$ for some $x \in [a,b]$ if $f,g$ continuous, $f(a) < g(a)$ and $f(b) > g(b)$ — Martin Sleziak, Apr 07 '17 at 14:41

score 2 · Accepted Answer · answered Apr 03 '17 at 15:50

2

Let $h=f-g$, $h(a)<0, h(b)>0$, the intermediate value theorem implies that there exists $x\in [a,b]$ with $h(x)=0$.

answered Apr 03 '17 at 15:50

Tsemo Aristide

87,475

Beautiful, I get it. Thank you much!! – liveFreeOrπHard Apr 03 '17 at 15:58

Suppose f and g are continuous on [a, b] and that f(a) < g(a), but f(b) > g(b). Prove that f(x) = g(x) for some x in [a, b].

1 Answers1