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 $g(c)=0$.
