Find values of $\alpha$ for $ f:\left[0,1\right]\rightarrow\left[0,1\right] $ such that $ f\left(c\right)=\alpha\cdot c$ I'm having a bit of trouble with a homework question. Here it is:
Let there be a function that $ f:\left[0,1\right]\rightarrow\left[0,1]\right]
 $ a continuous function.
For what values of $\alpha \in \mathbb{R}$ will there be a $c \in \left[0,1\right] $ such that $ f\left(c\right)=\alpha\cdot c$
I just can't fathom how I'm supposed to come up with values for $\alpha $ when the function is not defined for me (as in I don't have an equation for it).
Any hint?
Thanks
 A: A simple real analysis appoach is sufficient for this.However, I think a more graph-like approach is much better for your develop.
The Approach:

Image $f$ as a curve $y=f(x)$ , that  curve is in $Oxy$ and must sasitfy these:
i) $f$ connects 2 point in line $x=0$ and $x=1$ respectively 
ii)$f$ lies entirely in the cell $[0;1]\times[0;1]$
iii) (Not necessary for this problem) each line $x=c(0=c=1)$ cuts $f$ at exactly 1 point.
 Thus, the question of problem is to find which $\alpha$ so that $y=\alpha x$ and $f$ intersect forall all curve $f$ sastify i) and ii).
(Try to draw yourself some lines and curves , to know clearlier  )
We find that the line $y=\alpha x$ devides the cell in two parts .
Thus, the question of problem occurs if and only if  in that cell,line $x=1$ and $x=0$ lies in two different parts( not include endpoints) which are devided by y=\alpha x  
(Again, try draw some, it's good for anyone's knowledge)
Which equivalent to $\alpha \ge 1 $ .


In case, you want an alternative proof:
Another proach
For $\alpha <1$ , we choose the function $f_1(x)=1$, thus we easily see that $f_1(x)>\alpha alpha \forall x \in [0;1]$.Hence, these $\alpha$ do not sasity the conclusion.

For $\alpha \ge 1$ ,  $f(x) \in [0;1]$, therefore $ [ f(0)-\alpha.0][f( 1) -\alpha.1] \le 0$.
Due to the continuousness of $ f(x)-\alpha x$, we imply the conclusion.
