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# Root Hunter

$\mathbf{f(x) =\frac{1}{x-2} + \frac{1}{x-3}}$

For the argument to be valid, we must ensure that our functions are continuous in the range in question. For example, $f(1.9) \approx -10.9$ and $f(2.1) \approx 8.9$, but there is no zero between these values since the function in not continuous at $x = 2$.

However $f(2.9) \approx -8.9$, and $f$ is continuous for $2 < x < 3$ so there is a root between $2.1$ and $2.9$.

$\mathbf{f(x) = x^x - 1.5x}$

$f(1) =1^1 - 1.5 = -0.5$

and $f(2) = 2^2 - 1.5\times 2 = 4 - 3 = 1$

so there is a root between $1$ and $2$.

$\mathbf{f(x) = x^{1000000} + 1000000^x - 17}$

$f(-1) = (-1)^{1000000} + 1000000^{-1} - 17 = 1 + 10^{-6} - 17 = -15.999999$

and $f(1) = 1^{1000000} + 1000000^1 - 17 = 1 + 1000000 - 17 = 999984$

so there is a root between $-1$ and $1$.

$\mathbf{f(x) = \cos{(\sin{(\cos{x})})} -\sin{(\cos{(\sin{x})})}}$

$f(0) = \cos{(\sin{1})} - \sin{(\cos{0})} = \cos{0.84147 \cdots} -\sin{1} = 0.66637\cdots - 0.84147\cdots = -0.1751\cdots$

and $f(\frac{\pi}{2}) = \cos{(\sin{0})} - \sin{(\cos{1})} = \cos{0} - \sin{0.54030\cdots} = 1 - 0.5144\cdots = 0.4856\cdots$

So there is a root between $0$ and $\frac{\pi}{2}$.

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$\mathbf{f(x) =\frac{1}{x-2} + \frac{1}{x-3}}$

For the argument to be valid, we must ensure that our functions are continuous in the range in question. For example, $f(1.9) \approx -10.9$ and $f(2.1) \approx 8.9$, but there is no zero between these values since the function in not continuous at $x = 2$.

However $f(2.9) \approx -8.9$, and $f$ is continuous for $2 < x < 3$ so there is a root between $2.1$ and $2.9$.

$\mathbf{f(x) = x^x - 1.5x}$

$f(1) =1^1 - 1.5 = -0.5$

and $f(2) = 2^2 - 1.5\times 2 = 4 - 3 = 1$

so there is a root between $1$ and $2$.

$\mathbf{f(x) = x^{1000000} + 1000000^x - 17}$

$f(-1) = (-1)^{1000000} + 1000000^{-1} - 17 = 1 + 10^{-6} - 17 = -15.999999$

and $f(1) = 1^{1000000} + 1000000^1 - 17 = 1 + 1000000 - 17 = 999984$

so there is a root between $-1$ and $1$.

$\mathbf{f(x) = \cos{(\sin{(\cos{x})})} -\sin{(\cos{(\sin{x})})}}$

$f(0) = \cos{(\sin{1})} - \sin{(\cos{0})} = \cos{0.84147 \cdots} -\sin{1} = 0.66637\cdots - 0.84147\cdots = -0.1751\cdots$

and $f(\frac{\pi}{2}) = \cos{(\sin{0})} - \sin{(\cos{1})} = \cos{0} - \sin{0.54030\cdots} = 1 - 0.5144\cdots = 0.4856\cdots$

So there is a root between $0$ and $\frac{\pi}{2}$.

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?