Question

Let `f(x) = x^2 + 6` and `g(x) = (x + 8)/(x)` find `(g*f)(-7)`?

Answer 1

`-55/7`

Explanation:

Let's start by multiplying our functions, `f` and `g`. We are essentially setting up the following equation

`(x^2+6)/1=(x+8)/x`

Where we are essentially only multiplying the numerator to get

`(color(blue)((x^2+6)(x+8)))/x`

We can use FOIL (Firsts, Outsides, Insides, Lasts) to multiply this binomial. We get

`(x^2(x)+x^2(8)+6(x)+6(8))/x`

Which is equal to

`(x^3+8x^2+6x+48)/x` as our new function. We can plug in `-7` to get

`((-7)^3+8(-7)^2+6(-7)+48)/-7`

Which simplifies to

`(-343+392-42+48)/-7`

`=>-55/7`

Answer 2

`(g * f)(-7) = -55/7`

`(g @ f)(-7) = 63/55`

Explanation:

Given:

`f(x) = x^2+6`

`g(x) = (x+8)/x`

I suspect that the question is wanting the value of `(g @ f)(-7)` and not `(g * f)(-7)`, but let us describe both...

Case `(g * f)(-7)`

Given two functions `g` and `f` of a single variable, we can define the product function `g * f` by:

`(g * f)(x) = g(x) * f(x)" "` for all `x` in their common domain

With this definition:

`(g * f)(-7) = g(-7) * f(-7)`

`color(white)((g * f)(-7)) = ((-7)+8)/(-7) * ((-7)^2 + 6)`

`color(white)((g * f)(-7)) = (-1/7) * (49 + 6)`

`color(white)((g * f)(-7)) = -55/7`

Case `(g @ f)(-7)`

Given two functions `g` and `f` of a single variable, we can define their compostion `g @ f` by:

`(g @ f)(x) = g(f(x))color(white)(0/0)`
for all `x` in the domain of `f(x)` such that `f(x)` is in the domain of `g(x)`

With this definition:

`(g @ f)(-7) = g(f(-7))`

`color(white)((g @ f)(-7)) = g((-7)^2 + 6)`

`color(white)((g @ f)(-7)) = g(49 + 6)`

`color(white)((g @ f)(-7)) = g(55)`

`color(white)((g @ f)(-7)) = ((55)+8)/((55))`

`color(white)((g @ f)(-7)) = 63/55`