Question

Having a brain fart with domain and range again.. Just getting throw off because the x squared and x. Please little help. Dont have to give the answer, just how to get it.
Its. Square root of x^2-3x

Answer 1

\[x^2-3x \ge0\]

Answer 2

take the argument of the function, what is inside the radical and set in an inequality where the argument has to be greater of equal to zero.
\[x^2-3x \ge0\]

Answer 3

solver for x and then you will get what x cannot equal to.
so your domain would be all reals except what ever x is equal to.

Answer 4

I got that far giggles
but now not sure because the 2 xs

Answer 5

domain is real number R because any polynomila its domain would be real number inless it's mentioned

Answer 6

Would it be
x(x-3)>0. so X can not be less then or equal to 3?

Answer 7

But no because can be negative..

Answer 8

factor, find the zeros, graph it will be obvious where it is greater than 0

Answer 9

I have to solve with out graphing. Range and domain

Answer 10

{x ge\[x \ge 3\]

Answer 11

for the range , this funcction represent a parabola with a vertex
(-b/a, f(-b/a)) = (3/2,-9/4) ===> the range =[-9/4, +infinity)

Answer 12

i didn't mean graph the function itself, i meant graph
\[y=x^2-3x=x(x-3)\] get a parabola that opens up
positive outside the zeros, negative between them

Answer 13

Wont the domain be X>3 and X<0?

Answer 14

the function is
\[f(x)=\sqrt{x^2-3x}\] right? so you have to make sure
\[x^2-3x\geq 0\]

Answer 15

yes that is the domain you have it

Answer 16

Yes satellite. But how do i get that dominate without graphing? My equation only gets >3

Answer 17

\[x\leq 0\text{ or } x\geq 3\]

Answer 18

you mean the inequality
\[x(x-3)\geq 0\]?

Answer 19

domain not dominate sorry

Answer 20

without graphing you say
"if x > 3 both factors are positive, so their product is positive and
if x < 0 both factors are negative so their product is positive

Answer 21

I get that inequality. But when i solve it how am I finding the x<0

Answer 22

what i wrote, although the parabola in your head should get it as well. you can just reason it out. there is no need for algebra

Answer 23

if
\[x<0\] than both x and (x-3) are negative so their product is positive

Answer 24

The problem is my professor wants the algebra.. I know how to get it graphing obviously

Answer 25

well it really isn't algebra is it? you can make a little sign chart if you like

Answer 26

x ------------- 0 ++++++++++++++++++
x-3 -------------------------- 3 ++++++++++
x(x-3) +++++++0 ---------- 3 ++++++++++

Answer 27

As I stated above domain is clearly the whole real number set R because our function is a polynomial. and the range is (-9/4,+infinity)

Answer 28

kind of childish though don't you think? you already know that a parabola that faces up is positive outside the zeros and negative between them

Answer 29

@raheen it is
\[f(x)=\sqrt{x^2-3x}\] not
\[f(x)=x^2-3x\]

Answer 30

btw range is
\[[0,\infty)\] because square root symbol means the positive root

Answer 31

oh . I missed the square root notation, I'm sorry.

Answer 32

you are talking about what been posted but I'm solving for different data satellite

Answer 33

I dont think think thats the range. When i graph it, doesnt hit 0

Answer 34

domain is Real Number except (0,3)

Answer 35

sure it does.
\[f(0)=0 \] right?

Answer 36

You mean 0 to 3?

Answer 37

in fact
\[f(3)=0\] as well

Answer 38

domain
\[(-\infty,0]\cup [3,\infty)\]
range
\[[0,\infty)\]

Answer 39

Alright. Just wish I knew how to solve without graphing

Answer 40

hold on. what are you trying to solve "without graphing" the range?