Question

Find the range of the function f(x) /- 12x+24

Answer 1

Technically, it's \[f(x)\sqrt{12x+24}\]

Answer 2

Possible answers are...

A: \[y \le 12\]
B: \[y \ge 24\]
C: \[y \le 0 \]
D: \[y \ge 0\]

Answer 3

hmm have you covered parabolas yet?

Answer 4

This is radicals, but I've covered parabolas to some extent.

Answer 5

Why?

Answer 6

one sec

Answer 7

ok.... so.... the range
the range means, the "y values", or the values, "y" does take, based on whatever "x" is

so \(\bf f(x)=y=\sqrt{12x+24}\implies y^2=12x+24
\\ \quad \\
y^2-0x-0^2=12x+24\implies (y-0)^2=12x+24
\\ \quad \\
(y-0)^2=12(x+2)\impliedby \textit{"y" is squared, meaning a horizontal parabola}\)

any ideas on what the vertex for that parabola is?

where does it open towards? left or right?

Answer 8

I think it's supposed to open to the left?

Answer 9

the parabola is horizontal
has a vertex, or U-turn

"p" is positive, that is , the common factor for the "x", or 12, is positive, that means, opens to the right

Answer 10

Ah, so it opens to the right then, so what happens next?

Answer 11

hmm

Answer 12

lemme graph it.... somehow I think the rang is just from positive infinitiy to negative infinite

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJzcXJ0KDEyeCsyNCkiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjAsImVxIjoiLXNxcnQoMTJ4KzI0KSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi0xOC40MzY0MjU3ODEyNDk5OSIsIjIxLjIzNjQyNTc4MTI0OTk4OCIsIi0xMi40MDcwMzEyNDk5OTk5OTQiLCIxMi4wMDcwMzEyNDk5OTk5OTUiXX1d

Answer 13

the "domain" is restricted, since it's inside a root
so, it cannot make the root value negative, so "x" cannot be, as you can see in the graph
less than -2, otherwise the radicand turns negative, and the root becomes imaginary

but that's the domain

Answer 14

So if it's from positive infinity to negative infinity, it could be C or D?

Answer 15

the range... if you notice, just keeps on upwards and downwards

Answer 16

So it could be C? Equal to or less than zero?

Answer 17

well.... I think none of those choices are.... fit

since any of them are correct one can say, since the "range" as opposed to the domain
has a set of \(\large y \in \mathbb{R}\)

Answer 18

So which one would be the best choice?

Answer 19

hmm

Answer 20

Ah nuts, I just realized the thing was written the way it was, but with a = between fx and the rest of the problem. So it's the equation I wrote above, but with an = between Fx and the rest of it. Sorry.

Answer 21

Hopefully it won't impact it too much.

Answer 22

same thing... same equation really, yes, I noticed that \(\bf f(x)=y=\sqrt{12x+24}\implies y^2=12x+24
\\ \quad \\
y^2-0x-0^2=12x+24\implies (y-0)^2=12x+24
\\ \quad \\
(y-0)^2=12(x+2)\impliedby \textit{"y" is squared, meaning a horizontal parabola}\)

Answer 23

Oh good. Thank you. I was worried for a minute that you were going to get mad at me for wasting your time. So it's still the same thing?

Answer 24

nahh.... but.... hmmm the range is that "y" goes from \(\large (+\infty, -\infty)\)

Answer 25

So that means it could be one or the other?

Answer 26

hmmm lemme decompose it further.... I think it has a constraint

Answer 27

\(\bf f(x)=y=\sqrt{12x+24}\implies y^2=12x+24
\\ \quad \\
y^2-24=12x\implies \cfrac{y^2-24}{12}=x\)

notice that

based on that, I don't see any constraints on "y" per se

it could be negative anything, or positive anything
the square exponential, will make any negaative positive, or any positive positive anyway

Answer 28

So it could be pretty much anything?

Answer 29

yeap... notice the graph, the parabola just keeps on going upwards and downwards
slowly but surely

the one with the constraints is mainly the "domain", or the "x" value

Answer 30

So what's your opinion on what would be the best choice?

Answer 31

you could try reposting .... or tag someone else... maybe they have a different take on it

but I'm looking at the graph, and the range is just infinity to infinity

is it y>0 or y<12 or y>24 or whatever else, yes yes and yes

Answer 32

So I'm stuck one way or another?

Answer 33

I'll go with the 0 one then.

Answer 34

hhehe, sure, but basically, any choice will do

Answer 35

Yeah, I suppose if you state the answer goes off into infinity, might as well go with something that kind of comes close to it.

Answer 36

Thanks.

Answer 37

yw