Answer
a) $x^2+4x-5$
b) $x^2-4x+5$
c) $4x^3-5x^2$
d) $\dfrac{x^2}{4x-5}$
Domain: $\left(-\infty,\dfrac{5}{4}\right)\cup\left(\dfrac{5}{4},\infty\right)$
Work Step by Step
We are given the functions:
$f(x)=x^2$
$g(x)=4x-5$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=x^2+4x-5$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=x^2-(4x-5)=x^2-4x+5$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=x^2(4x-5)=4x^3-5x^2$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{x^2}{4x-5}$
The domain of $\dfrac{f}{g}$ is the set of all real numbers except the zeros of $g$:
$4x-5=0\Rightarrow x=\dfrac{5}{4}$
The domain is:
$\left(-\infty,\dfrac{5}{4}\right)\cup\left(\dfrac{5}{4},\infty\right)$
