f(x)=4x-1
g(x) = 3-2x
^{2}
h(x)=sq rt (x+5)
h(g(2k+1))
Domain and Range of the composite function, and the composite function itself.
I went at it this way;
h(3-2(2k+1)
^{2}
h(3-2(4k
^{2}+4k+1)
3-8k
^{2}-8k-2
h(-8k
^{2}-8k+1)
Substitute it into sq rt (x+5), giving us;
sq rt (-8k
^{2}-8k+6)
Now I dont know if thats correct, or if I should divide all figures by 2, leaving us with;
Why divide by 2?
sq rt(-4k
^{2}-4k+3)
and then the square root sign and exponent of 2 cancel out?
You have an expression that contains a square, not a squared expression. Nothing to cancel.
For domain, I don't know how to figure that out, I would assume you don't carry the original domain of h(x) which was x> or equal to -5
and range I would assume would have to be greater than or equal to 3 since that is the point at the parabola's vertex
That is not the vertex. How did you come up with that? $- \dfrac{-8}{2(-8)} = -\dfrac{1}{2}.$
$6 - 8\left(- \dfrac{1}{2}\right) - 8\left(- \dfrac{1}{2}\right)^2 = 6 + 4 - 2 = 8.$
Now what?
Any help is appreciated everyone, this is my final unit of this course and I owe a lot of it to the helpful people on this forum, so thanks again!