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Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

How Many Solutions?

Find all the solutions to the this equation.

After Thought

Which is larger cos(sin x) or sin(cos x) ? Does this depend on x ?

Curve Match

Age 16 to 18 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 $a \neq 1$