Answer
a) $x^2+3x-15$
b) $-x^2+3x+17$
c) $3x^3+x^2-48x-16$
d) $\dfrac{3x+1}{x^2-16}$
Domain: $\left(-\infty,-4\right)\cup\left(-4,4\right)\cup\left(4,\infty\right)$
Work Step by Step
We are given the functions:
$f(x)=3x+1$
$g(x)=x^2-16$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=3x+1+x^2-16=x^2+3x-15$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=3x+1-(x^2-16)=3x+1-x^2+16=-x^2+3x+17$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=(3x+1)(x^2-16)=3x^3+x^2-48x-16$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{3x+1}{x^2-16}$
The domain of $\dfrac{f}{g}$ is the set of all real numbers except the zeros of $g$:
$x^2-16=0\Rightarrow x^2=16\Rightarrow x=\pm 4$
The domain is:
$\left(-\infty,-4\right)\cup\left(-4,4\right)\cup\left(4,\infty\right)$
