# Thread: Existance of Interval in continuous function

1. ## Existance of Interval in continuous function

Let f(x) be a continuous function on [0, 3] such that f(0) = f(3). Show that there exists an element, c, in [0, 2] such that f(c) = f(c+1).

2. Originally Posted by kjwill1776
Let f(x) be a continuous function on [0, 3] such that f(0) = f(3). Show that there exists an element, c, in [0, 2] such that f(c) = f(c+1).
Consider the continuous function $g(x) = f(x + 1) - f(x)$ on the interval $\left[ {0,2} \right]$.
Consider: $\left. \begin{gathered} g(0) = f(1) - f(0) \hfill \\ g(1) = f(2) - f(1) \hfill \\ g(2) = f(3) - f(2) \hfill \\ \end{gathered} \right\}\; \Rightarrow \;g(0) + g(1) + g(2) = 0$.

If each of $g(0),g(1),g(2)$ is zero then we are done.

Else, this means that some one of the pairs, $\left\{ {g(0),g(1)} \right\},\left\{ {g(2),g(1)} \right\},\left\{ {g(0),g(2)} \right\}$ has one negative and one positive member. WHY?

Now apply the intermediate value theorem to that pair.

3. Originally Posted by Plato
Else, this means that some one of the pairs, $\left\{ {g(0),g(1)} \right\},\left\{ {g(2),g(1)} \right\},\left\{ {g(0),g(2)} \right\}$ has one negative and one positive member. WHY?

Now apply the intermediate value theorem to that pair.
I'm struggling to find the pair to which I need to apply the IVT. Could you elaborate on this idea, specifically why do we know one pair has one negative and one positive number?

4. Originally Posted by kjwill1776
I'm struggling to find the pair to which I need to apply the IVT. Could you elaborate on this idea, specifically why do we know one pair has one negative and one positive number?
The point is we cannot be sure which pair has one positive and one negative member.
We just know one must exist. Say it is the pair $\{g(1),g(2)\}$.
That means that zero is between them so $\left( {\exists c \in (1,2)} \right)\left[ {g(c) = 0} \right]$ which inturn implies that $f(c + 1) = f(c)$.