Continuous function problem

**tttcomrader** · November 13th 2006, 01:59 PM

Q: Suppose that f: [0,2] -> [0,4] is continuous. Show that there is an x $\epsilon$ [0,2] such that $f(x) = 2x$ .

Proof: All I know is use the intermediate value theorem...

**Plato** · November 13th 2006, 03:26 PM

What happens if f is nowhere differentiable?
Are you saying that f is a surjection (onto)?
There has got to be more to this question than you posted!

**tttcomrader** · November 13th 2006, 04:23 PM

That is the whole problem. But there is a hint here provided by my professor:

Hint: Consider the function g: [0,2] -> R defined by g(x) = f(x) - 2x and use the intermediate value theorem.

Now, again, we have not learn about using derivatives in proof yet.

Thank you!

**ThePerfectHacker** · November 13th 2006, 05:00 PM

Define a function on the interval $[0,2]$ as,
$g(x)=f(x)-2x$
Since, $f(x),2x$ are countinous on $[0,2]$ so too is $g(x)=f(x)-2x$ continous on $[0,2]$ .
Now,
$g(0)=f(0)-0=f(0)$
$g(2)=f(2)-4$

By definition $f(0)\geq 0$

If $f(0)=0$ then the proof is complete.
Because $x=0\in [0,1]$ is one such point. So there is no harm in assuming that $f(0)>0$

By definition $f(2)\leq 4$

If $f(2)=4$ then the proof is complete.
Because $x=2\in [0,2]$ is one such point. So there is no harm in assuming that $f(2)<4$

Thus, $g(0)=f(0)>0$ and $g(2)=f(2)-4<0$ .

Since $g(x)$ is continous on $[0,2]$ there is a point $c\in [0,2]$ such that $g(c)=0$ by the intermediate value theorem.
Thus, $g(c)=f(c)-2c=0$ . Thus, $f(c)=2c$ for some point in $[0,2]$ .
Q.E.D.

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