How do I find the Domain of
x+4 / square root of 3x+1?
Square root of 2x+5?
X / X(X-3)?
I have a few more questions on the bottom. Thanks
correct
remember, the domain of a function is the set of all inputs (in this case, x-values) for which the function is defined. We see that the function is undefined if the denominator is zero, so we must have $\displaystyle \sqrt {3x + 1} \neq 0$. But there is also a restriction on the square root, what is being square rooted cannot be negative. The only way we can satisfy both these conditions is if $\displaystyle 3x + 1 > 0$. Thus the domain is all x such that $\displaystyle 3x + 1 > 0$ or, more formally,
(Using set notation) $\displaystyle dom(f) = \{ x \mbox { : } 3x + 1 > 0 \} = \{ x \mbox { : } x > 1/3 \}$
(Using interval notation) $\displaystyle dom(f) = \left( - \frac {1}{3} , \infty \right)$
Thanks guys this helped a bunch~ I'm currently learning operations and compositions of functions and I have a few questions I was wondering you could answer for me.
1. f(x) = x^2, g(x) = squareroot 1-x
I'm trying to find (f o g)(x) and it's domain. The problem i'm having is I know that (f o g)(x) = 1 - x so shouldn't the Domain be (-Infinite, Infinite)?
2. Similarily same functions but finding (g o f)(x). It is = Squareroot 1 - x^2. So to find the Domain, I should use 1 - x^2 greater than equal to 0?
In both these cases i'm having trouble understanding what the Domain should be. If anyone can help explain it would be great.
Oh and for one of the above questions I had earlier, Squareroot 2x + 7 domain should be 2x+7 Greater than equal to 0 where it should end up x greater than equal to [-7/2, infinite) right?
recall my explanation of what the domain is in one of my previous posts in this thread. it is the set of all x's for which the function is defined.
you are correct in that $\displaystyle (g \circ f)(x) = \sqrt {1 - x^2}$
and you are correct that the domain is all x such that $\displaystyle 1 - x^2 \geq 0$ since the square root is only defined for these values.
now to solve, which i don't see what your problem is. we have the difference of two squares on the left
$\displaystyle 1 - x^2 \geq 0$
$\displaystyle \Rightarrow (1 + x)(1 - x) \geq 0$
$\displaystyle \Rightarrow -1 \leq x \leq 1$
thus, the domain is $\displaystyle [-1,1]$
you typed $\displaystyle \sqrt {2x + 5}$ before. which is it? anyway, the domain for the function you found above is correct. except you don't say "x greater than equal to [-7/2, infinite)" that statement makes no sense! it is x >= -7/2 or dom(f) = [-7/2, infinity)