Question

Let f(x) = (x^2 + 3x - 4) and g(x) = (x+4). Find f/g and state the domain

Answer 1

Also, I multiplied the two of these and got x^3 + 8x^2 - x - 6. Is that right?

Answer 2

You can multiply f and g together, but that's not the point of this particular problem. Instead, you are to DIVIDE f(x) = (x^2 + 3x - 4) by g(x) = (x+4) and then state the domain of the resulting quotient.

\[\frac{ x^2+3x-4 }{ x+4.}\]

Answer 3

Before you do anything else, notice that x cannot = -4. Why is that?

Next, try factoring the numerator.

Can one of the factors of the numerator cancel out the denominator?

Please share y our work.

Answer 4

Oh, part a was to multiply the two of them. I thought it'd be easier to get my work checked for that too in this question than making another question lol

Answer 5

x cannot equal -4 because there is a domain restraint? I know the domain factored is (x - 1)(x + 4)^2. The first of the factors might be able to cancel out the denominator... But I'm not sure. Can you use half of a factorization to do that? Isn't that like taking half of a number?

Answer 6

I did most of the long division already. I get tripped up on the part after I divide out the (x+4) into the x^2. What do I do for the 3x spot?

Answer 7

Note that the numerator factors to (x+4)(x-1). If the denominator is (x+4), you may cancel (x+4) from numerator and denominator, leaving (x-1). But note that even though cancelling is possible, this does not remove the restriction that x cannot = -4.

How would you write the domain of this function?

Answer 8

Um D: \[-4 < x \le \infty\]?

Answer 9

Or is it all real numbers except -4?

Answer 10

Yes: all real numbers except -4. Very good!

Answer 11

Oh thanks

Answer 12

so you have combined two functions f and g through division and identified the domain of the new function thus obtained. ;)

Answer 13

x - 1 is a function? Does that mean all expressions are functions?