Question

f(x)=sq rt (4-7x) can some one please explain how to find the range and domian of this equation

Answer 1

domain will be the values "x" can take
range will be the values "y" can take, keep in mind that "y" values depend on whatever value "x" turns into

Answer 2

so the domian wi be (-infiny,+infinty) since its doesnt equal zero?

Answer 3

so as far as roots with even radical, that is square root in this case
restrictions inside radical pertain to the issue that the radical amount inside CANNOT BE NEGATIVE

so any values "x" will consider in taking, will have to take that into consideration

so, what values can "x" take without making the radical negative?
that is, what "x" can be without making 4-7x a negative figure?

Answer 4

well, let's see on that, let's use the value x = 1
4-7x => 4-7(1) => -3
well, that's negative and \(\bf \sqrt{-3} \implies i\sqrt{3}\) which is an imaginary number
so 1 for "x" is not good, since it makes the radical value negative

Answer 5

oh okay give me a second to figure it out and ask you if its right

Answer 6

notice that \(\bf \sqrt{0} = 0\) so that's ok, it's just when you get into the negative when the issue turns to imaginary

Answer 7

Okay so since you start of at zero and can positive its (-infiy,0]

Answer 8

can't*

Answer 9

hmm, lemme give you a few values there
\(\bf f(x)=\sqrt{4-7x}\\
x= 1, \qquad \sqrt{4-7(1)} \implies \sqrt{-3}\qquad \textit{no good}\\
x= 0, \qquad \sqrt{4-7(0)} \implies \sqrt{4}\qquad \textit{good}\\
x = \cfrac{4}{7} \qquad \sqrt{4-7\left(\frac{4}{7} \right)} \implies \sqrt{0}\qquad \textit{good}\\
\textit{if "x" goes above }\cfrac{4}{7} \textit{then you get negative values for the radical}\)

Answer 10

so the values "x" can safely take without making the radical negative are, any negative value up to \(\bf \cfrac{4}{7} \implies (-\infty, \cfrac{4}{7}]\)

Answer 11

So this would be the range correct? and this only accounts for problems that have square roots.

Answer 12

the range consequently due to that restriction to "x", will be that yes

Answer 13

as far as "x", x can take anything

Answer 14

this restriction accounts to roots that have an EVEN RADICAL
that is \(\bf \Large \sqrt[4]{x}\qquad \sqrt[6]{x}\qquad \sqrt[20]{x} ...\)

Answer 15

on an ODD RADICAL that doesn't happen, an example of that will be \(\bf \sqrt[3]{-27} \implies \sqrt[3]{-3\times -3 \times -3} \implies \sqrt[3]{(-3)^3}\implies -3\)

Answer 16

a negative number can have an ODD root, thus it doesn't happen with ODD RADICALS, just with EVEN ones

Answer 17

Okay so the range is (-infy,4/7] and the domain is [0,infy)

Answer 18

the domain will be \(\bf (-\infty, \quad +\infty)\)

Answer 19

so, "x" can take anything, but "y" will be restricted to only the positive values in the radical

Answer 20

hmmm, not sure if I'm confusing... maybe I'm ... heheh, the domain and range,

Answer 21

Im a bit confused im trying to learn this out of pauls note introduction ex 4 letter b and it explains it differently

Answer 22

I think I got them backwards, the restrictions to "x" are to the domain

Answer 23

as far as the range, well, oddly enough because an even root can spit out negative and positive roots, will be \(\bf (-\infty, \quad +\infty)\)

Answer 24

why, well, because \(\Large y = \pm \sqrt{4-7x}\)

so any value the radical becomes, the root itself will give a positive and negative account of it

Answer 25

am I making any sense? hehe

Answer 26

Give me a sec im read everything that you explained and he has and let you know if i get it. lol

Answer 27

\(\bf y = \pm \sqrt{4-7x}\\
x = -100 \qquad y = \pm \sqrt{74} \implies \pm 8.6\\
x = -1,000,000 \qquad y = \pm \sqrt{1,000,004} \implies \pm 1000.002\)

Answer 28

do you do tuyoring by any chance?

Answer 29

so the \(\bf \Large \pm \) part of the root, will give all real numbers

Answer 30

could, but no

Answer 31

I under stand now. Thanks a bunch. Have a great day

Answer 32

http://www.mesacc.edu/~scotz47781/mat120/notes/radicals/domain/square_root/square_roots.html