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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- Sep 8th 2009, 03:40 PMMorgan82Domain and Range
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. - Sep 8th 2009, 04:02 PMpickslides
$\displaystyle x \in (3,\infty)$

$\displaystyle y \in \mathbb{R} /{0}$ - Sep 8th 2009, 04:09 PMluobo
- Sep 8th 2009, 05:57 PMpickslides
Given the range of the function is the domain of the inverse then y can be negative

$\displaystyle f(x)= \frac{2}{\sqrt{x-3}}$

$\displaystyle f^{-1}(x) = \frac{4}{x^2}+3$