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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 $f(x)=y$?

Pedro
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sequence
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  • have you tried drawing? – L F Feb 03 '16 at 01:05
  • Yes. But under any point on the curve of $f(x)=2x(1-x)$, there is a corresponding $x$, such that $f(x) = \text{a point on the curve}$. Isn't it true? The curve is continuous, and no points are missing there. While I do realize that the interval for the range of $f(x)$ is shorter, both ranges are uncountable intervals. – sequence Feb 03 '16 at 01:09
  • what about $0.8$ ? is there any point $x$ in $[0,1]$ such $f(x)=0.8$? – L F Feb 03 '16 at 01:10
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    Since it's a portion of an upside-down parabola with zeros at $x = 0$ and $x = 1$, its maximum occurs at the average of the two zeros, at $x = 1/2$. Just plug $x = 1/2$ in and see if your function is tall enough to hit everything between $0$ and $1$. – pjs36 Feb 03 '16 at 01:10
  • @LuisFelipe: $0.8$ is not in the range of $f(x)$. – sequence Feb 03 '16 at 01:11
  • @sequence you are asking why $g(x)$ is not onto in $[0,1]$ – L F Feb 03 '16 at 01:17

4 Answers4

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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 $2x(1-x)=1$.

Brian M. Scott
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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.

Pedro
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  • Now it's becoming clear :) – sequence Feb 03 '16 at 01:16
  • @sequence To think about the concept of "onto" it's not enough consider only the range; you can't forget the codomain. The function is onto if the range equals the codomain. – Pedro Feb 03 '16 at 01:21
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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

image

L F
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  • Maybe I'm confused with a definition. I thought that a surjective function is such a function which, for any $y$ in the range of $f$, there is an $x$, such that $f(x)=y$ is in the range of $x$. Maybe I'm confusing something. – sequence Feb 03 '16 at 01:12
  • in this example, $f$ is surjective because all points in $[0,1]\subset Y$ has a preimage. But $g\colon[0,1]\to[0,1]$ is not surjective because the points above of $0.5$ doesnt belongs to range of $g$ – L F Feb 03 '16 at 01:19
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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.

siegehalver
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