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What are the extreme values of the function $$y = (x^2 - 1)^{1/2} = \sqrt{x^2 - 1}$$ and where do they occur?

I have gotten as far as finding that $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}}$$ What next? Set $dy/dx$ to $0$?

Null
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Ethan
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    Yes, set the derivative to 0 and solve for $x$. Then plug $x$ back into the equation for $y$. – Null Oct 28 '14 at 21:36
  • The issue with that is i'm getting x=0, and if I plug that in then y = i, and I know that's not the answers. I have the answers in the book, and imaginary numbers definitely aren't there. – Ethan Oct 28 '14 at 21:38
  • Check your derivative, I don't think its correct – Pakquebchsoflwty Oct 28 '14 at 21:39
  • Also, if the derivative is undefined, then it could be an extreme value as well. – Pakquebchsoflwty Oct 28 '14 at 21:40
  • Could someone correct me on what i'm doing wrong? I set dy/dx to 0, so the equation then reads 0 = x/(sqrt(x^2-1)), so I multiply both sides by the denominator and end up with x=0. – Ethan Oct 28 '14 at 21:42
  • If complex numbers aren't allowed then the domain of the function does not include $|x| \leq 1$. – Null Oct 28 '14 at 21:48
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    When x=-1,1, then it could be an extreme value since dy/dx is undefined. – Pakquebchsoflwty Oct 28 '14 at 21:53
  • I would say, the investigation function is always necessary to start the domain of the function. It is not possible to derive a function outside its domain. – georg Oct 28 '14 at 22:08

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Whenever you want to know something about a function's values, you first need to know what the domain of the function even is. In this case, the function is defined for $|x|\ge1$, that is, on $(-\infty,-1] \cup [1,\infty)$.

A wonderful property of continuous functions is that extreme values (local maxima and minima) can occur only at critical points and at endpoints of the domain. To find the critical points, you find everywhere in the domain that the derivative either vanishes or is undefined. Here, the derivative doesn't vanish anywhere in the domain; it is undefined at $x=\pm1$. The endpoints of the domain are at $x=\pm1$ anyway. So the only possible local extrema are at $x=-1$ and at $x=1$. And sure enough, those are local minima (the values there are $0$ and the function is positive elsewhere).

If by "extreme values" you mean global minima and maxima, then you need to check $\lim_{x\to\infty} y(x)$ and $\lim_{x\to-\infty} y(x)$ as well, since the domain extends without end in both directions. In this case, you'll find that the function has no global maximum, and $0$ is the global minimum.

Greg Martin
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