How do I prove the maximum of this function I have the function
$$y = x - \sqrt{x^2 - 1}$$
which must have a maximum of $1$ at $x = 1$, as after that you're taking $x$ and subtracting something slightly smaller than $x$, tending to $0$ as $x$ tends to infinity, however its derivative of 
$$1 - \frac{x}{\sqrt{x^2 - 1}}$$
is undefined at $x = \pm 1$, as is its second derivative. 
How can I prove this function is bounded above by 1, and that the absolute value of y doesn't exceed 1 at some point 0 < x < 1?
 A: If you want to use the derivative, note that $\frac{x}{\sqrt{x^2-1}}\gt 1$ for $x\gt 1$. Thus $\frac{dy}{dx}\lt 0$ for $x\gt 1$, and therefore our function is increasing in the interval $(1,\infty)$. 
Remark: Your non-calculus argument was fine. Maybe it is clearer to multiply top and bottom by $x+\sqrt{x^2-1}$. We find that
$$x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}.\tag{1}$$
It is clear that as $x$ increases from $1$, the right side of (1) decreases.
A: Enough to consider $x\ge 0$. Make the substitution $x= \cosh t$, $\sqrt{x^2-1}= \sinh t$ with $t \ge 0$.
$$x- \sqrt{x^2-1} = \cosh t - \sinh t = e^{-t}$$  with maximum $1$ at $t=0$ 
A: HINT: If you consider one-sided derivatives, you can see, that $y$ is monotonous on $(1-\delta,1)$ and (the other type of monotonicity) on $(1,1+\delta)$.
A: To get maximum of this function, clearly, $x \ge 0$. So we have:
$$x-\sqrt{x^2-1} = x-\sqrt{x^2-1} \times \frac{x+\sqrt{x^2-1} }{x+\sqrt{x^2-1} } = \frac{1 }{x+\sqrt{x^2-1} } \ge \frac{1}{1+\sqrt{1^2-0}}= 1,$$
Note that $x+\sqrt{x^2-1}$ is an increasing function at $[1, +\infty]$.
A: Consider  the total picture, that of the hyperbola. In many elementary situations the square root indicates that another branch exists side by side.
$$ y^2 - 2 x y +1 =0 $$
Two roots are component curve branches 
$ y_1 = x + \sqrt {x^2-1} $  and 
$ y_2 = x - \sqrt {x^2-1} $ 
Graph the same after interchanging axes. For rotated axes minimum point is at 1,1) and maximum point at -1,-1) which are quite regular. 
If you deal with it as it is, it will create some confusion.
I shall post the graphs
A: Another way is that for $x > 1$, note that
$$ x-x\sqrt{1-\frac1{x^2}} < x-x\left(1-\frac1{x^2}\right)=\frac1x < 1$$
