Homework

sorry, the indicated operation is f/g. That equation is about as far as I got.

>>327426
You gotta factorise bro, I'll give it a go but it's been like 10 years.

what operations?
also domain is ]-inf, inf[ \ {1 ± 4}

What I need to do it find f/g.

In the picture I put the f function in the numerator and the g function in the denominator

factorisation: x * (2x - 5) / ((x-5)(x+3))

what do you mean by f/g? i don't see a way the further simplify it

>>327438
I don't really see any way to simplify either, thats one of the answers but I just wanted to make sure. Also I mean the function f divided by the function g.

>>327424
is it supposed to be f(x) = 2 * (x^2-5x) ? then you could factorize it to 2 * x *(x-5) and could simplify f/g to 2x / (x+3)

>>327429
How did you get the domain? it seems to me like it could just be all real numbers

File: Untitled.png (3KB, 164x71px) Image search: [Google]

3KB, 164x71px

This is as simple as it gets, unless the question is wrong

>>327444

the function just is not defined for x = -3 or x = 5

>>327444
>>327446
the denominator is zero for x=-3 and x=5

>>327447
I just tested it an you're right, however I'm still a little stumped as to how you got those answers?

>>327451
you know how you're not supposed to divide by 0?
if x = -3 or x = 5 then the denominator is 0 and you end up dividing by 0 which is not a valid operation

have you checked if the numerator is supposed to be 2 * (x^2 -5x) as suggested in >>327443

>>327443
no I'm positive the numerator in the OP is correct.

>>327453
gotcha, so If I were to write this in interval notation it would be (-inf, -3)U(5,inf) correct?

no in interval notation it would be ]-inf, -3[ U ]-3,5[ U ]5, inf[
the way you wrote it down you're missing all the possible values between -3 and 5 which are possible, including the roots at x=0 and x = 2.5

>>327459
hot damn, thanks anon. I've got some studying to do.

>>327461

how old are you? or rather, what class are you in?

>>327463
I'm a bit ashamed but I'm in precalculus algebra in college. I just haven't taken a math class in 6 years.

>>327465
what are you majoring in?

anyways I'm bored, if you've got more homework, I'd be happy to help

>>327468
dual major in computer security systems and network management, just an AAS though.

I actually do have another problem, pretty much because I'm garbage at multiplying fractions with variables.

>>327473
I know for f(x), it cannot be anything less than 4. and for g(x) it cannot be 8. I'm not sure how to write it down in interval notation or if I'm missing something due to multiplying the functions.

>>327473

to find the domain, you basically always ask yourself: "what would break the function?", i.e. what values do I have to insert for x
to divide by 0 or
to have the square root of a negative number or
to have the logarithm of a number smaller or equal to zero?

>>327473
x is any number greater than or equal than 4, except 8.

>>327474

there sometimes might be a case where you can continuously proceed(I'm not sure about the proper english terminology) the function. for example if you have f(x) = (x-5)(x-3) and g(x) = x-3 and you want the domain of f/g, and then it depends on your TA's definition of domain

otherwise the domain of the combined function should be the combination of your domains, in your case ]4, 8[ U ]8, inf[

>>327474
>>327479
sorry [4,8[ ]8,inf[

>>327481
Thanks man

File: homew.png (713B, 141x34px) Image search: [Google]

713B, 141x34px

The question for this one is to determine whether it is even, odd, or neither.

I went with neither since changing the x to -x didn't give me the exact same equation nor did it give me the opposite. I would like to make sure though.

>>327484

for polynomials you can use as a rule of thumb that if there are only even exponents, the function is even, if there are only odd exponents the function is odd.
the constant term is an "even" term as you could write it as your constant * x^0

tl;dr you're right