Question

Find (f/g) (x) for the functions provided: ƒ(x) = x3 − 27, g(x) = 3x − 9

Answer 1

Answer:

$(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)$

Step-by-step explanation:

We have been given that

$f(x)=x^3-27,g(x)=3x-9$

We can use the formula for difference of cubes to simplify the function f(x)

difference of cubes -  $a^3-b^3=(a-b)(a^2+ab+b^2)$

$f(x)=x^3-27\\\\=x^3-3^3\\\\=(x-3)(x^2+3x+9)$

And g(x) can be written as

$g(x)=3x-9\\=3(x-3)$

Thus, we have

$(\frac{f}{g})(x)=\frac{(x-3)(x^2+3x+9)}{3(x-3}$

On cancelling the common factors, we get

$(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)$