Question

find (a) (f*g)(x) and the domain of f*g.
f(x)=x^2-3x g(x)= sqrt(x+2)
answers: (f*g)= x+2-3 sqrt(x+3)
domain: [-2,infinity)
I How did the square root cancel out in the first part of the answer? did it cancel out when it was squared?
How do you find the domain? Any help is great! Thank you, all!

Answer 1

\( f(x) = x^2-3x \\ g(x) = \sqrt {x+2} \)
right ?
and do you need to find \((f*g)(x) \: or \: (f \: o\: g)(x)\) ??

Answer 2

since, square root of a negative number is not real, we have to remove all the values of x for which the quantity inside square root becomes negative.
so, we must include values of x which makes \(x+2 \ge 0\)
right ?
\(x+2 \ge 0 \implies x \ge -2\)
which when written in interval notation is \([-2, \infty ) \)
do you know why those type of brackets "[" and ")" are put there ?

Answer 3

Yes you are correct. and (fog)(x) is what I need to find. I had the correct idea but with the answer " x+2-3 sqrt(x+2)" I was not sure how the "x+2" in the first part of the answer did not have a square root anymore, but the other half did.
As for the "[" and ")", you told me that the "[]" had to do with being included with in the domain and that it can take particular value. And the "()" are not included in the domain. I also remember that parentheses' are always used for infinities.

Answer 4

good :)
so (fog) (x)
you can find fog (x) by substituting 'x' in f(x) by g(x)

Answer 5

like example if f(x) = x^3 , g(x) = log x
then fog(x) = (log x)^3
note that i have replaced 'x' in f(x) by log x
which was g(x)

Answer 6

can you find fog (x) for your question ?

Answer 7

also note \((\sqrt y)^2=y\)

Answer 8

Yes, I can find fog. I was just unsure of the (y√)^2=y part. Other than that I was sure of everything else, and I was a little shaky on the "find the domain" part. But overall, I did remember how to find fog, I just did not know what to do when it came to squaring a square root lol So thank you, very very much for your help!

Answer 9

welcome, very very much ^_^