Apply the intermediate value theorem to the intersection of two functions

Question

Prove the following using the intermediate value theorem:

$f$, $g$ are continuous on $[a,b]$, $f(a)< g(a)$ but $f(b) >g(b)$

Prove: $f(c)=g(c)$ for some $c \in (a,b)$

I am not sure if i am correct:

$g$ is continuous: using IVT $g(c)=j$, $g(a)< j <g(b)$ , c belongs to (a,b)
- $f$ is continuous: using IVT $f(c)=k$, $f(a)< k < f(b)$ , therefore by the given inequality: $f(a)< j <f(b)$ Hence, by IVT $j= f(c)= g(c)$

g is continuous: using IVT g(c)=l g(a)<j<g(b j belongs to (a,b) f is continuous: using IVT f(c)=k f(a)< k < f(b) , k belongs to (a,b)
Therefore by the given inequality: f(a)< j <f(b) Hence,by IVT j= f(c)= g(c) Is this way correct? — user146656, May 06 '14 at 16:39
No: it doesn't seem to be correct: where do you think you proved the existence of a point where both functions $;f,g;$ have the same value? — DonAntonio, May 06 '14 at 16:40
By the given inequality it can be shown that j lies between f(a) and f(b) , and since it lies between them by the intermediate value theorem it can be shown that j= f(c) ? — user146656, May 06 '14 at 16:44
I really can't see it clrealy, @user...did you try the idea in my first comment? — DonAntonio, May 06 '14 at 16:45
if we can show that j lies between f(a) and f(b) is it correct to conclude that j= f(C)? — user146656, May 06 '14 at 16:49
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 15:04

score 2 · Answer 1 · answered May 06 '14 at 16:49

2

Following my coment:

$$h(x):=f(x)-g(x)\implies \begin{cases}h(a)=f(a)-g(a)<0\\{}\\ h(b)=f(b)-g(b)>0\end{cases}$$

Since $\;h\;$ is continuous, the IVT tells us there exists...(complete)

answered May 06 '14 at 16:49

DonAntonio

@user146656, you begin your coment with "using IVT g(c)=l g(a)<j<g(b j belongs to (a,b)" . I can't see here how "j" relates to "c" or to "l" ...and you should really try to learn a basic MathJax to write properly mathematics in this site: that'll help a lot to the understanding. Try here:http://meta.matheducators.stackexchange.com/questions/93/mathjax-basic-tutorial-and-quick-reference – DonAntonio May 06 '14 at 17:02
I have edited the question. That was just a typing mistake. – user146656 May 06 '14 at 17:08
1

Do you think it is, @user146656 ? I really can't even make much sense of it...how miraculously do we get $;j=f(c);$?? This would mean $;k=j;$ , according to your notation... – DonAntonio May 06 '14 at 17:24

1 Answers1