If $f$ and $g$ are continuous on $[a,b]$, $f(a) \le g(a)$, and $g(b) \le f(b)$, prove there is a $c$ in $[a,b]$ with $f(c)$ $=$ $g(c)$.

Question

If $f$ and $g$ are continuous on $[a,b]$, $f(a)$ $\le$ $g(a)$, and $g(b)$ $\le$ $f(b)$, prove there is a point $c$ $\in [a,b]$ such that $f(c)$ $=$ $g(c)$.

Any ideas on how to solve?

I think I have to use the Intermediate Value Thm but I'm not sure.

You probably meant $g(b)\le f(b)$. In this case, consider the function $f-g$. — David Mitra, Apr 15 '14 at 12:01
See also: 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 15:05

5xum · Accepted Answer · 2014-04-15T12:19:55.260

3

Hint: Try to construct a function $h$ from $f$ and $g$ with the properties: $$h(a) \leq 0$$ $$h(b) \geq 0$$ $$h(x)=0\iff f(x)=g(x)$$

Then, what does the intermediate value theorem tell you about $h$?

edited Apr 15 '14 at 12:19

answered Apr 15 '14 at 12:04

5xum

123,496
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"$h(x) = 0 \iff f(a) = f(b)$"?! I don't understand what you're trying to say there. – kahen Apr 15 '14 at 12:19
@kahen Me neither. I was writting in a hurry. I fixed it now. – 5xum Apr 15 '14 at 12:20

score 0 · Answer 2 · edited Apr 16 '17 at 06:40

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Hint If you can't use the intermediate value theorem, try to find $\sup_x f(x)-g(x) >0$ and apply continuity there.

edited Apr 16 '17 at 06:40

Martin Sleziak

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answered Apr 15 '14 at 12:16

Sylvain Biehler

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If $f$ and $g$ are continuous on $[a,b]$, $f(a) \le g(a)$, and $g(b) \le f(b)$, prove there is a $c$ in $[a,b]$ with $f(c)$ $=$ $g(c)$.

2 Answers2