Challenge Level

We really liked this solution sent in from Alex from Stoke on Trent Sixth Form College:

The horizontal axis is the $x$-axis, and the vertical axis the $y$-axis.

I used $x=0.5$ to calculate the y values and ordered them.

The curves, from top to bottom on the graph, are the functions arranged in decreasing order of $y$ value for any $x$ value between $0$ and $1$ when they are equal.

The order becomes $2\sqrt{x}-x\,,\sqrt{x}\,, x\,, (x^2+x)/2\,, x^2$

Any function of the form $x^n, n> 2$ will return smaller $y$ values than $x^2$ between $0$ and $1$; avoiding the existing curves and intersect at the end points. $x^{\frac{1}{n}}\,, n> 2$ is similar, instead returning values higher than $x^{\frac{1}{2}}$.

Functions of this form also meet the conditions. To generate a curve $c(x)$ such that $a(x) < c(x) < b(x)$ for $0 < x < 1$, and $a(0) = b(0) = c(0), a(1) = b(1) = c(1)$, the arithmetic mean could be used. $c(x) = (a(x)+b(x))/2$, or a more general $c(x) = a(x) + k(b(x)-a(x)), 0 < k < 1.$ $k = 0.5$ becomes the arithmetic mean. This formula can be used with $a(x)$ and $b(x)$ as adjacent existing curves (such as $a(x) = x, b(x) = \sqrt{x}$). The existing functions $\sqrt{x}$ and $(x^2+x)/2$ could be thought of as having been generating by using $2\sqrt{x}-x$ and $x$, and $x$ and $x^2$, respectively. $a(x) = 0, b(x) = x^2$, and $a(x) = 2\sqrt{x}-x$, $b(x) = 1$ produces $c(x)$ curves below and above existing curves.

Resulting $c(x)$ functions from using the general formulae (or the specific arithmetic mean case) can be used as the $a(x)$ or $b(x)$ functions in generating others, resulting in many different functions involving $x, \sqrt{x}$ and $x^2$ terms.

However, there are other functions in addition that meet the conditions, such as the binary logarithm. $y = \ln(x+1)/\ln(2) = \log_2(x+1)$ Between $0$ and $1$, $x < \log _2(x+1) < \sqrt{x}$, therefore it does not intersect the curves except for the end points where: $\log_2(0+1) = 0$ and $\log_2(1+1) = 1$.

Simon said

The $x$ axis is horizontal and the
$y$ axis is vertical.

Royal blue $y = 2\sqrt{x} -x$

Dark orange $y = \sqrt{x}$

Lime green $y = x$

Turquoise $y =
\frac{x^2+x}{2}$

Purple $y = x^2$

There are an infinite number of curves that can be found that
will intersect only at the two end points.

One potential form is $y = \frac{x^2+ax}{1+a}$ for $a \neq
1$