# Example of how to use intermediate value theorem when given the range of c.

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 $f(c)=\lambda$.

Question:

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:

$$f(c)=2c^2$$

Attempt:

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?

Hint:

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 $g(c)=0$.

• Why do we consider this, and where do we go to next? – UniStuffz May 21 '16 at 18:34
• You have to prove there exists $c\in(0,1)$ such that $g(c)=0$. So compute $g(0=$ and $g(1)$. – Bernard May 21 '16 at 18:43
• Ok, and what do we define $f(x)$ to be? – UniStuffz May 21 '16 at 18:57
• You can't compute g(x) without knowing f(x) right? – Piotr Benedysiuk May 21 '16 at 19:00
• @UniStuffz the inequalities you have for $f(0),f(1)$ are enough. – 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 $f(c) = 2c^2$.

• Why do you think such $x_0,x_1$ exist? – almagest May 21 '16 at 18:24
• 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. – Piotr Benedysiuk May 21 '16 at 18:33
• We are not told it is surjective. The result is true even if it is not. – almagest May 21 '16 at 18:35
• 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. – Piotr Benedysiuk May 21 '16 at 18:59
• Yes you can use IVT for the case $f(x)=1$ for all $x$ (but you don't apply it to the function $f(x)$ ). – almagest May 21 '16 at 19:00