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

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 $\frac{\pi}{2}$.