# implied domain

• Feb 9th 2008, 09:10 PM
RoboStar
implied domain
Hey MHF: Ive just got a question on the implied domain.
f(x)= Square root(x^2 - 3)

easy question i just need an somebody to clarify how to solve it.

Cheers
Andrew
• Feb 9th 2008, 09:16 PM
Jhevon
Quote:

Originally Posted by RoboStar
Hey MHF: Ive just got a question on the implied domain.
f(x)= Square root(x^2 - 3)

easy question i just need an somebody to clarify how to solve it.

Cheers
Andrew

the domain of $\displaystyle \sqrt{x}$ is $\displaystyle x \ge 0$ for $\displaystyle x \in \mathbb{R}$. thus, the implied domain here is given by: $\displaystyle x^2 - 3 \ge 0$
• Feb 9th 2008, 09:33 PM
RoboStar
and another question

h(x) = square root(x - 4) + square root(11 - x)
• Feb 9th 2008, 09:40 PM
Jhevon
Quote:

Originally Posted by RoboStar
and another question

h(x) = square root(x - 4) + square root(11 - x)

you just have to make sure $\displaystyle x - 4 \ge 0$ and $\displaystyle 11 - x \ge 0$ work at the same time. choose the values of x that satisfy both
• Feb 10th 2008, 01:27 AM
RoboStar
thanks for all your help just two more questions on implied domain

f (x) = x^2 - 1 / x + 1 and h (x) = square root(x-1/x+2)
• Feb 10th 2008, 07:40 AM
Jhevon
Quote:

Originally Posted by RoboStar
thanks for all your help just two more questions on implied domain

f (x) = x^2 - 1 / x + 1 and h (x) = square root(x-1/x+2)

okay, for rational functions, we have the condition that we cannot divide by zero. and you know the condition for square roots. so there are three conditions you want to fulfill here

for f(x): you want $\displaystyle x + 1 \ne 0$

for h(x): you want $\displaystyle \frac {x - 1}{x + 2} \ge 0$ (for the square root) and $\displaystyle x + 2 \ne 0$ (for the rational function being square rooted)

of course, for h(x), both conditions must be fulfilled at the same time. so find the x's that work for both simultaneously