Question

f(x) = 3x + 2; g(x) = 3x - 5

Find f/g.

Answer 1

\[\large\rm \color{royalblue}{f(x)=3x+2},\qquad \color{orangered}{g(x)=3x-5}\]
Then,\[\large\rm \frac{\color{royalblue}{f(x)}}{\color{orangered}{g(x)}}\]just plug in the functions, ya? :)

Answer 2

3x+2/3x-5?

Answer 3

then what would the domain be?

Answer 4

Yes,
(3x+2)/(3x-5)

The domain is generally going to be all real numbers.
But since we have a fraction, we might have a restriction at a location.

Answer 5

Recall that in the land of math, we can not divide by 0.
\(\large\rm \frac{1}{0}\) <-- This is not a number, this is undefined. This is bad.

So what we'll do is, we'll `let` our denominator equal 0 like this.
And see if we can find these "bad" values.

Answer 6

\[\large\rm 3x-5=0\]This should give us the "bad" value for x.

Answer 7

5/3?

Answer 8

Good good good.
This is the bad value.

Which means our domain consists of every x `except` x=5/3

Answer 9

so would the full answer be (f/g)(x) = Quantity three x plus two divided by three x minus five.; domain {x|x ≠ Five over three. }

Answer 10

Can I ask one more?

Answer 11

so would the full answer be (f/g)(x) = Quantity three x plus two divided by `the quantity` three x minus five.; domain {x|x ≠ Five over three. }


I would probably add this in, so it's clear that the bottom is an entire quantity, not just a 3x denominator.

Answer 12

Sure

Answer 13

f(x) = 7x + 7, g(x) = 6x2

Find (fg)(x).

Answer 14

ok thanks:)

Answer 15

\[\large\rm \color{royalblue}{f(x) = 7x + 7},\qquad\qquad \color{orangered}{g(x) = 6x^2}\]

\[\large\rm (fg)(x)=\color{royalblue}{f(x)}\color{orangered}{g(x)}\]\[\large\rm =\color{royalblue}{(7x+7)}\color{orangered}{6x^2}\]So here I guess you'll have to do some distributing.

Answer 16

so 42x^3+42x?

Answer 17

?

Answer 18

42x^3+42x^2 ya?