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 $f(c)=g(c)$.

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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:



  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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!



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 $h$.


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, $c = a = b$

  • $\begingroup$ 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

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