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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?

Qiri
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6 Answers6

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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.

André Nicolas
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  • But my argument and these answers show that it decreases from x =1 up, but how can I show that the modulus of the answer if x is between -1 and 1 is less than 1? – Qiri Nov 18 '14 at 14:27
  • There is no answer between $-1$ and $1$, the function is not defined, for there $x^2-1$ is negative, has no real square root. – André Nicolas Nov 18 '14 at 14:29
  • I mean the maximum of the norm, sorry I should have said. – Qiri Nov 18 '14 at 14:30
  • I've edited my question to show what I left out. – Qiri Nov 18 '14 at 14:32
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    The function is not defined in $(-1,1)$. If you wish to go to complex numbers, we need to choose the principal branch of the square root, take the norm. I don't think it throws useful light on our real variables problem. – André Nicolas Nov 18 '14 at 14:38
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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 $t=0$

orangeskid
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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 $(1,1+\delta)$.

Przemysław Scherwentke
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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 $[1, +\infty]$.

Paul
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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

Narasimham
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Another way is that for $x > 1$, note that
$$ x-x\sqrt{1-\frac1{x^2}} < x-x\left(1-\frac1{x^2}\right)=\frac1x < 1$$

Macavity
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