# Math Help - Composition of Functions

1. ## Composition of Functions

f(x)=4x-1
g(x) = 3-2x2
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(4k2+4k+1)
3-8k2-8k-2

h(-8k2-8k+1)
Substitute it into sq rt (x+5), giving us;

sq rt (-8k2-8k+6)

Now I dont know if thats correct, or if I should divide all figures by 2, leaving us with;

sq rt(-4k2-4k+3)

and then the square root sign and exponent of 2 cancel out?

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

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!

Any input?

3. ## Re: Composition of Functions

Originally Posted by rhickman
f(x)=4x-1
g(x) = 3-2x2
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(4k2+4k+1)
3-8k2-8k-2

h(-8k2-8k+1)
Substitute it into sq rt (x+5), giving us;

sq rt (-8k2-8k+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(-4k2-4k+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!
The problem is $h(g(2k + 1))\ given h(x) = \sqrt{x + 5}\ and\ g(x) = 3 - 2x^2.$

$g(2k + 1) = 3 - 2(2k + 1)^2 = 3 - 2(4k^2 + 4k + 1) = 3 - 8k^2 - 8k - 2 = 1 - 8k - 8k^2.$ This was fine.

$h(1 - 8k - 8k^2) = \sqrt{1 - 8k - 8k^2 + 5} = \sqrt{6 - 8k - 8k^2}.$ Good job.

Now for what values of k is that a real-valued function?

$6 - 8k - 8k^2 = 0 \implies k = \dfrac{8 \pm \sqrt{(-8)^2 - 4(-8)(6)}}{2 * (-8)} = \dfrac{8 \pm \sqrt{64 + 192}}{-16} = \dfrac{8 \pm 16}{-16} = \dfrac{1}{2}\ or \dfrac{-3}{2}.$

Now what in terms of domain?

4. ## Re: Composition of Functions

The question only asks for the domain, just wanted to mention that correction firstly.

Secondly, by graphing the function it shows that the functions roots are at the numbers mentioned above from the quadratic equation, and that those are the final values for x, so x must be greater than or equal to -1.5 and less than or equal to .5

I think that's right, again i double checked with the graph.

Thanks for more input today.

5. ## Re: Composition of Functions

Originally Posted by rhickman
The question only asks for the domain, just wanted to mention that correction firstly.

Secondly, by graphing the function it shows that the functions roots are at the numbers mentioned above from the quadratic equation, and that those are the final values for x, so x must be greater than or equal to -1.5 and less than or equal to .5

That is correct. Using the graph is fine. You can also use test points. Finally, you can use qualitative reasoning: as the absolute value of k becomes large, $-8 k^2$ is going dominate the expression and make it negative.

I think that's right, again i double checked with the graph.

Thanks for more input today.
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