Theorem: Let $f$ be continuous on $[a,\,b]$ and assume $f(a)\not=f(b)$. Then for every $\lambda$ such that $f(a)<\lambda<f(b)$, there exists a $c\in(a,\,b)$ such that $f(c)=\lambda$.


Suppose that $f : [0,1] \rightarrow [0,2]$ is continuous. Use the Intermediate Value Theorem to prove that their exists $c \in [0,1]$ such that:



I know that when we have the condition were $f : [a,b] \rightarrow [a,b]$, the method to prove that c exits, is the same method you would use to prove the fixed point theorem.

Unfortunately I don't have an example in my notes when we have $f : [a,b] \rightarrow [a,y]$. How would I use the IVT to answer the original question?



Consider the function $g(x)=f(x)-2x^2$.

Some details:

$g(0)=f(0)\ge 0$ since the range of $f$ is contained in $[0,2]$. Similarly, $g(1)=f(1)-2\le 0$. Furthermore, $g$ is continuous since $f$ is. Now, either $g(0)$ or $g(1)==0$, and there's nothing to prove. Or $g(0)>0$, $g(1)<0$. The Intermediate value theorem assures there exists $c\in (0,1)$ such that $g(c)=0$.

| cite | improve this answer | |
  • $\begingroup$ Why do we consider this, and where do we go to next? $\endgroup$ – UniStuffz May 21 '16 at 18:34
  • $\begingroup$ You have to prove there exists $c\in(0,1)$ such that $g(c)=0$. So compute $g(0=$ and $g(1)$. $\endgroup$ – Bernard May 21 '16 at 18:43
  • $\begingroup$ Ok, and what do we define $f(x)$ to be? $\endgroup$ – UniStuffz May 21 '16 at 18:57
  • $\begingroup$ You can't compute g(x) without knowing f(x) right? $\endgroup$ – Piotr Benedysiuk May 21 '16 at 19:00
  • $\begingroup$ @UniStuffz the inequalities you have for $f(0),f(1)$ are enough. $\endgroup$ – almagest May 21 '16 at 19:02

First note: $\exists x_0, x_1$ such that $f(x_0) = 0$, $f(x_1) = 2$. Both x are somewhere in the interval $[0,1]$.

Second note: $0 < 2c^2 < 2$ for $ c \in (0,1)$.

So we have: f is continious on $[x_0, x_1] \subset [0, 1]$, $f(x_0) \neq f(x_1)$ and $f(x_0) < 2c^2 < f(x_1)$. By IVT there must exist $c \in (x_0, x_1)$ so that $f(c) = 2c^2$.

| cite | improve this answer | |
  • $\begingroup$ Why do you think such $x_0,x_1$ exist? $\endgroup$ – almagest May 21 '16 at 18:24
  • $\begingroup$ Fair enough, such x's don't need to exist. I'm wildly saying that f(x) assumes every value in [0,2] at least once. By no means is that true - f(x) = 0 is also a mapping from [0,1] to [0,2]. Lets say f assumes more than one value on [0,1]. Call those a,b. We can now generalize my result by exchanging 0 and 2 in the first line by a and b. $\endgroup$ – Piotr Benedysiuk May 21 '16 at 18:33
  • $\begingroup$ We are not told it is surjective. The result is true even if it is not. $\endgroup$ – almagest May 21 '16 at 18:35
  • $\begingroup$ Say f(x) = 1 for all x. Then we can't use ITV, which was specificly asked of us. I don't see a problem with my proof other than math version of grammar nazi. $\endgroup$ – Piotr Benedysiuk May 21 '16 at 18:59
  • $\begingroup$ Yes you can use IVT for the case $f(x)=1$ for all $x$ (but you don't apply it to the function $f(x)$ ). $\endgroup$ – almagest May 21 '16 at 19:00

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.