Had my first calc test today! I feel pretty good about it. One question... y = 2/ sqrt(x-3) find domain and range... Can someone tell me what it is... I hope I did it right.
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$\displaystyle x \in (3,\infty)$ $\displaystyle y \in \mathbb{R} /{0}$
Originally Posted by pickslides $\displaystyle x \in (3,\infty)$ $\displaystyle y \in \mathbb{R} /{0}$ Shouldn't $\displaystyle y $ be greater than zero?
Last edited by mr fantastic; Sep 18th 2009 at 08:11 AM. Reason: Restored original reply
Given the range of the function is the domain of the inverse then y can be negative $\displaystyle f(x)= \frac{2}{\sqrt{x-3}}$ $\displaystyle f^{-1}(x) = \frac{4}{x^2}+3$
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