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 $f(a)&lt;\lambda&lt;f(b)$, 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 $g(0)&gt;0$, $g(1)&lt;0$. The Intermediate value theorem assures there exists $c\in (0,1)$ such that $g(c)=0$.
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: $0 &lt; 2c^2 &lt; 2$ 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 $f(x_0) &lt; 2c^2 &lt; f(x_1)$. By IVT there must exist $c \in (x_0, x_1)$ so that $f(c) = 2c^2$.