$2x(1-x)$ is not onto? How come $4x(1-x)$ is onto in $[0,1]$ but $2x(1-x)$ is not? Isn't it true that for any $y$ in the range interval, there exist two $x$ such that $f(x)=y$?
 A: The graph of $y=2x(1-x)$ is a parabola opening down. It has its vertex halfway between the zeroes at $x=0$ and $x=1$, i.e., at $x=\frac12$, and $2\cdot\frac12\left(1-\frac12\right)$ is only $\frac12$. Thus, no value of the function is greater than $\frac12$, and it cannot map $\Bbb R$ onto $[0,1]$.
If you prefer an algebraic approach, note that 
$$2x(1-x)=2x-2x^2=-2(x^2-x)=-2\left(\left(x-\frac12\right)^2-\frac14\right)=\frac12-2\left(x-\frac12\right)^2$$
by completing the square. $2\left(x-\frac12\right)^2\ge 0$ for all $x$, so
$$\frac12-2\left(x-\frac12\right)^2\le\frac12$$
for all $x$. In particular, there is no value of $x$ such that $2x(1-x)=1$.
A: To get an answer, you have to specify the codomain.


*

*$f:[0,1]\to[0,1]$ given by $f(x)=4x(1-x)$ is onto;

*$f:[0,1]\to[0,1]$ given by $f(x)=2x(1-x)$ is not onto;

*$f:[0,1]\to[0,2]$ given by $f(x)=4x(1-x)$ is not onto;

*$f:[0,1]\to[0,1/2]$ given by $f(x)=2x(1-x)$ is onto.
A: It is enought to see the graph of both functions. $g(x)$ is not onto because not all elements of $[0,1]$ have a preimage 

A: Well $f(x)=-2x^2+2x$, so $f'(x)=-4x+2$. Thus we have our only critical point $x=\frac{1}{2}$, and $f(\frac{1}{2})=\frac{1}{2}$. Thus since $f$ is increasing on the interval $(-\infty,\frac{1}{n}] $ and decreasing on $[\frac{1}{n}, \infty)$, we have that $f$ has a maximum of $\frac{1}{2}$, and your result follows.
