Question

What is the range of f(x)=√(3x-7) -0.5?

I got [3,∞) which I believe is correct for the domain, but I'm not sure what the range would be and if it would be [-0.5,∞) or not

Answer 1

Or if it's [0.91, ∞)?

Answer 2

\(\bf f(x) = \sqrt{ (3x-7) } - 0.5 \quad ?\)

Answer 3

Yes that's the equation

Answer 4

well, with EVEN radicals, that is, like square root, or root 2 <-- even
is that you cannot have a negative value inside it, otherwise you'd end up with an "imaginary" value
so the value in it has to be positive

so any values "x" can take, MUST NOT make the radical quantity, 3x - 7, negative
0 is ok, -1 or less is not good, because \(\bf \sqrt{ 0 } = 0\) , so that's ok
so at what point will that be for "x"?

well, we can easily find out by setting the equation to 0
3x - 7 = 0
3x = 7
\(\bf x = \frac{ 3 }{ 7 }\)

so, the lowest "x" can go is \(\bf \frac{ 3 }{ 7 }\) , at that point, the radical gives 0
and y = -0.5

so "x" cannot go lower than \(\bf \frac{ 3 }{ 7 }\) , it can go higher though, up to \(\bf +\infty\)

Answer 5

ohh... I messed one thing.... \(\bf x = \frac{ 7 }{ 3 }\) rather

Answer 6

So the range would then be [7/3, ∞)?

Answer 7

yeap

Answer 8

Thank you!

Answer 9

yw

Answer 10

at that point, y = -0.5
from there, as the radical gives a positive value, then "y" goes upwards

Answer 11

well, a plus version and minus version will come out of the radical...so the range will ... go upwards and downwards

Answer 12

So, the domain is [3, ∞) and the range is [7/3, ∞). I think that sounds right

Answer 13

So the range would then be [7/3, ∞)? <--- ... I misread you.... that's correct for the DOMAIN

Answer 14

Ohhh.... okay, sorry, it got confusing

Answer 15

the values "x" takes, are the domain of the function

Answer 16

and [-0.5, ∞) is the range, thusly? Is that correct?

Answer 17

recall that \(\bf f(x) = \sqrt{ (3x-7) } - 0.5 \implies f(x) = \pm \sqrt{ (3x-7) } - 0.5\)
so you see, there's a + and a - version from the radical
so I'd say it goes to \(\boldsymbol{(-\infty, +\infty)}\)

Answer 18

But when x is 2 and below, there is no y value, so wouldn't that make the domain never be negative?

Answer 19

I'm afraid not, the EVEN roots, have 2 values from it, a + and a -, so any value resulting from a x = 5, 25, 1000....
will yield a +y and -y exact value

keep also in mind that \(\bf f(x) = \sqrt{ (3x-7) } - 0.5 \implies f(x) = \pm \sqrt{ (3x-7) } - 0.5\\ \quad \\ \quad \\
y = \sqrt{ (3x-7) } - \cfrac{ 1 }{ 2 } \implies y + \cfrac{ 1 }{ 2 }= \sqrt{ (3x-7) }\\ \implies \left(y + \cfrac{ 1 }{ 2 }\right)^2=3x-7\\\quad \\
\cfrac{ 1 }{ 3 }\left(y + \cfrac{ 1 }{ 2 }\right)^2+\cfrac{ 7 }{ 3 }=x\)

is a parabola

Answer 20

I'm a still a little confused but alright

Answer 21

this would be the graph of it => http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIrc3FydCgzeC03KS0xLzIiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjAsImVxIjoiLXNxcnQoM3gtNyktMS8yIiwiY29sb3IiOiIjMDAwMDAwIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTMuNCIsIjkuNiIsIi00LjAzOTk5OTk5OTk5OTk5OSIsIjMuOTYiXX1d

Answer 22

notice the 2 + and - roots

Answer 23

I only need the positive graph of it, though. Not the negative square root graph.

Answer 24

hmm... well... , I'd think you need both.... but maybe the exercise is only concerned with the positive one

Answer 25

yeah, it is only concerned with the positive one

Answer 26

ok... then your interval of \(\boldsymbol{ [-0.5, +\infty) }\) is correct

Answer 27

Oh, ok, good, thanks for all the help!

Answer 28

yw