3.
f ⋅ g = f(x) ⋅ g(x)
Since f(x) = -5x + 3 , we can replace f(x) with -5x + 3
f ⋅ g = (-5x + 3) ⋅ g(x)
Since g(x) = 6x - 2 , we can replace g(x) with 6x - 2
f ⋅ g = (-5x + 3) ⋅ (6x - 2)
Before we simplify, let's get the domain.
There are no denominators or anything, so the domain is all real numbers.
f ⋅ g = (-5x)(6x) + (-5x)(-2) + (3)(6x) + (3)(-2)
f ⋅ g = -30x2 + 10x + 18x + -6 Combine like terms...
f ⋅ g = -30x2 + 28x - 6
4.
\(\frac{f}{g}\,=\,\frac{f(x)}{g(x)} \\~\\ \frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)
Before simplifying, let's get the domain. x + 4 is in a denominator, so...
x + 4 ≠ 0
x ≠ -4 The domain is all real numbers with the exception that x ≠ -4
\(\frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)
Factor the numerator as a difference of squares.
\(\frac{f}{g}\,=\,\frac{(x-4)(x+4)}{x+4}\)
Reduce the fraction by (x + 4) .
\(\frac{f}{g}\,=\,\frac{(x-4)}{1} \\~\\ \frac{f}{g}\,=\,x-4\)