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

##### 2 Answers

#### Explanation:

Let's start by multiplying our functions,

Where we are essentially only multiplying the numerator to get

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

Which is equal to

Which simplifies to

#### Explanation:

Given:

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

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

I suspect that the question is wanting the value of

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

Given two functions

#(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 @ 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#