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?

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