5
$\begingroup$

First of all, let me write the statement properly:

Theorem :

Let $f(x)$ and $g(x)$ are continuous on a closed interval $[a,b]$. If $f(a)< g(a)$ and $f(b)>g(b)$, then there exists a $c$ in the interval $[a,b]$ such that $f(c)=g(c)$.

enter image description here

I am new at proofs, so I wanted ask if the proof below correct? I feel like Ive jumped logical step. If so please let me know, thanks.

Proof: If $f$ is continuous on $[a,b]$, then by the Intermediate Value Theorem, there exists a $c$ such that $f(c)= L$ where $L$ is between $f(a)$ and $f(b)$. And similarly, by the same theorem there exists a $c$ such that $g(c)= K$ where $K$ is between $g(a)$ and $g(b)$.

We must prove that there is at least one value in the interval such that $K=L$.

Since we know $f(a)< g(a)$, therefore at some point

$$ f(a)- g(a) < 0 \tag{1}\label{eqn1} $$

and at a distinct point in the interval we know the following will be true:

$$f(b)-g(b)>0 \tag{2}\label{eqn2} $$

Now define a function $h(c) = g(c)-f(c)$. It follows from (1) and (2) that $h(c)>0$ at one point and $h(c)<0$ at another point then by the intermediate value theorem at some point on the interval $h(c)$ must equal to zero. So,

$$h(c)= f(c)- g(c) = 0$$ Therefore, $$f(c)=g(c)$$

For reference to the Intermediate Value Theorem see the following links:

https://www.mathsisfun.com/algebra/intermediate-value-theorem.html

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&cad=rja&uact=8&ved=0CFsQFjAIahUKEwjdppOg66HHAhUFjw0KHUQvCOM&url=http%3A%2F%2Fwww.cut-the-knot.org%2FGeneralization%2Fivt.shtml&ei=3lbKVd2_H4WeNsTeoJgO&usg=AFQjCNE4cTobZvz7jim1bUQPx1A35H3jzw&sig2=rsKlZooQCf-Cw0_TWTqMTg&bvm=bv.99804247,d.eXY

$\endgroup$
  • $\begingroup$ Also for curiosity's sake does anyone know the name of the theorem? $\endgroup$ – Red Aug 6 '15 at 0:09
  • $\begingroup$ Intermediate value theorem $\endgroup$ – Shailesh Aug 6 '15 at 0:14
  • 1
    $\begingroup$ Just apply the IMVT to the function $h(x)=f(x)-g(x)$ $\endgroup$ – Mark Viola Aug 6 '15 at 0:14
  • $\begingroup$ How do you go from $f(b)-g(b)>0$ to a contradiction of $f(b)-g(b)=0$? That seems wrong to me. $\endgroup$ – JB King Aug 6 '15 at 0:21
  • 1
    $\begingroup$ once you realize the existence of $c$, you simply need to write $f(c)-g(c)=0$ and you are basically done. "So we get at $c$,..." you can basically delete everything after that and just say $f-g$ is continuous therefore there is a $c$ such that $f(c)-g(c)=0$. $\endgroup$ – jdods Aug 6 '15 at 0:22
1
$\begingroup$

You got off to a great start! Rather, since $f-g$ is continuous on $[a,b]$ (why?) and $$(f-g)(a)<0<(f-g)(b),$$ then at some point $c$ between $a$ and $b$, you know that $(f-g)(c)=0,$ meaning $f(c)=g(c),$ and you're done!

$\endgroup$
5
$\begingroup$

Hint:

1) The difference of two continuous functions is continuous;

2) Define $h(x) = g(x) - f(x)$. Then $h(a) > 0$ and $h(b) < 0$;

3) Apply the Intermediate Value Theorem to $h$.

$\endgroup$
3
$\begingroup$

Your proof seems a bit confused, all you need is that $f - g$ is positive at one end point and negative at the other, and thus must be $0$ somewhere in between because $f - g$ is continuous (Why?). I don't see how you get, for example, $c = a = b$

$\endgroup$
  • $\begingroup$ yeah I could clean up the language a bit. By I was applying the intermediate value theorem to the function (f-g). I should of wrote h(x)=(f-g) first I get. $\endgroup$ – Red Aug 6 '15 at 0:31
  • $\begingroup$ I got the idea in my head, I messed up in articulating it. $\endgroup$ – Red Aug 6 '15 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.