4. "I got the square root of 7x. Is this right?"
Nearly, but not quite. Careful with your sqrt signs.
f(x)=sqrt(x) g(x)=6sqrt(x)
f(x)+g(x) = sqrt(x) + 6 sqrt(x) = 7 sqrt(x), not sqrt(7x)
16. This is much easier on paper!
Consider your constants. You have sqrt(2) on top, and 2sqrt(2) on bottom. Cancel the sqrt(2)s to leave 2 on the bottom, or 1/2
Consider your variable. You have x^1/2 on top and x^1/3 on bottom. That gives you x^(1/6) - the sizth root of x.
So you end up with 1/2*x^1/6
29. I'm not sure why the domain (not range!) would not be just all real numbers. (The range will always be 0 or positive, of course)
4. Find f(x) + g(x) when f(x) = x^(1/2) and g(x) = 6x^(1/2).
===> I got the square root of 7x. Is this right?
16. Find f(x)/g(x) when f(x) = (2x)^(1/2) and g(x) = 2*(sqrt2)*x^(1/3).
===> I got (sqrt2x) / (3*cubertx), but I definitely don't think this is right....if it isn't, can someone show me how to do it please? :)
29. Find the domain of f(f(x)) when f(x) = x^2.
===> f(f(x)) would equal x^4, right? Weeeeeell, how would you graph this to find the domain? I thought it would just be all real numbers greater than zero, but that doesn't seem right -_-
Any help is GREATLY appreciated!!! :D
2 answers
#4 your wrote : "I got the square root of 7x. Is this right?"
If by that you mean 7√x, that would be correct, but the way you expressed it sounds like √(7x).
√(7x) would be wrong.
16. just take (2x)^(1/2) / 2√2x^(1/3)
= x^(1/2 - 1/3) / 2
= x^(1/6) / 2
29.
if f(x) = x^2 , then f(f(x)) = x^4
(you had that)
Now what kind of x's can you put in there ?
Wouldn't any x be allowed ?
So the domain is the set of real numbers.
If by that you mean 7√x, that would be correct, but the way you expressed it sounds like √(7x).
√(7x) would be wrong.
16. just take (2x)^(1/2) / 2√2x^(1/3)
= x^(1/2 - 1/3) / 2
= x^(1/6) / 2
29.
if f(x) = x^2 , then f(f(x)) = x^4
(you had that)
Now what kind of x's can you put in there ?
Wouldn't any x be allowed ?
So the domain is the set of real numbers.