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I don't get this question from my Calc textbook, can u please explain it to me.

How do you get from $\cfrac{\sqrt{x-3}}{x-3}$ to $\cfrac {1}{\sqrt{x-3}}$?

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    Multiply the top and bottom of the original expression by $\sqrt{x - 3}$2012-10-31

4 Answers 4

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Devide top and bottom by $\sqrt{x-3}$

$\frac{\sqrt{x-3}}{x-3}=\frac{(x-3)^{1/2}\cdot (x-3)^{-1/2}}{(x-3)\cdot(x-3)^{-1/2}}=\frac{1}{\sqrt{x-3}}$

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$\forall\;\; 0< x\in\Bbb R\;\;,\;\;\frac{\sqrt x}{x}=\frac{1}{\sqrt x}$

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    I guess. I think \forall x >0, x\in\Bbb R is clearer. Or $\forall x\in \Bbb R^+$2012-10-31
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Note that $\frac1a=a^{-1}$; that for $a>0$ we have $a^x\times a^y=a^{x+y}$; and lastly $\sqrt a=a^{\frac12}$.

Apply that for $a=x-3$ in the case at hand:

$\frac{\sqrt a}a=\frac{a^{\frac12}}{a}=a^{\frac12}\times a^{-1}=a^{\frac12-1}=a^{-\frac12}=\frac1{a^{\frac12}}=\frac1{\sqrt a}$

That is to say: $\frac{\sqrt{x-3}}{x-3}=\frac1{\sqrt{x-3}}.$

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$\frac{1}{\sqrt{x-3}}=\frac{1}{\sqrt{x-3}}\cdot \frac{\sqrt{x-3}}{\sqrt{x-3}}=\frac{\sqrt{x-3}}{|x-3|}=\frac{\sqrt{x-3}}{x-3},$ when $x>3$, which is assumed since the domain is

$\lbrace x\in \mathbb{R}:x>3\rbrace $.