# Functions and their domains

• Oct 30th 2007, 10:09 AM
Coach
Functions and their domains
I was looking over my notes recently, and this kind of problem was written
$f(x)=\sqrt{5-3x}$

The domain is defined for all values of x<(inclusive) $\frac{5}{3}$, right?

then I have something weird written for function
$\frac{1}{\sqrt{5-3x}}$

the domain is defined for x belongs to $]-\infty,\frac{5}{3}]\cup$x belongs to $[\frac {5}{3},\infty[$

This can't be right?

I clearly rememver myself saying that the answer is x is greater than 5 over three, and the teacher said it was correct.
• Oct 30th 2007, 10:16 AM
SnipedYou
Yes it is $x \le \frac{5}{3}$

This is because nothing in the radical can be negative so 0 has to be the lowest we can go, so $\frac{5}{3}$ is the lowest you can go before going into imaginary numbers, which cannot be graphed on the real plane.

For the second one you have written x cannot be 0 so the domain would be $x < \frac{5}{3}$