Question

Please help, I need someone to talk me through the steps...

Determine the domain of the function: 3x÷x(x^2−25)

Answer 1

Sounds like you're dealing with
\[\Large \frac{3x}{x(x^2-25)}\]
Is my assumption correct?

Answer 2

Yes, your assumption is correct. @jim_thompson5910

Answer 3

Focus strictly on the denominator: \(\Large x(x^2-25)\)

Answer 4

Set it equal to zero and solve for x. Tell me what you get

Answer 5

Alright I'm stuck on \[x^3=25\]

Answer 6

Did you use the idea that if A*B = 0, then A = 0 or B = 0 ?

Answer 7

x(x^2-25) = 0
x = 0 or x^2-25 = 0

so we know that x = 0. That's one of the there values. There are 2 more

Solve
x^2 - 25 = 0
for x

Answer 8

Ohhhh alright one sec

Answer 9

So \[x=\sqrt{25}\]

Answer 10

what is the square root of 25 equal to?

Answer 11

Oh 5 duh.. Sorry, it's been a long day in pre-cal :P

Answer 12

don't forget about the plus minus

eg:

x^2 = 100
x = 10 or x = -10

Answer 13

Ugh.. Okay so x=0, x=5, x=-5?

Answer 14

So x = 0 or x = 5 or x = -5 will make the denominator equal to zero. Therefore, we must kick these values out of the domain. Any other real number will be a valid input.

Answer 15

Ohhh I see, so the domain is all real numbers except for those three!

Answer 16

Correct. In `set-builder notation` the domain would be
\[\Large \left\{x| x\in\mathbb{R}, \ x \ne 0, \ x \ne 5, \ x \ne -5\right\}\]

In `interval notation`, the domain would be
\[\Large (-\infty, -5) \cup (-5, 0) \cup (0, 5)\cup(5,\infty)\]
The idea here is that we start with (-infty, infty) and we poke holes at the values where we're pulling out of the domain

Answer 17

Got it! I think I understand now! :)

Answer 18

I'm glad it makes more sense