Answer
a) $x-3$
b) $3x-7$
c) $-2x^2+9x-10$
d) $\dfrac{2x-5}{2-x}$
Domain: $(-\infty,2)\cup(2,\infty)$
Work Step by Step
We are given the functions:
$f(x)=2x-5$
$g(x)=2-x$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=2x-5+2-x=x-3$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=2x-5-(2-x)=2x-5-2+x=3x-7$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=(2x-5)(2-x)=4x-2x^2-10+5x=-2x^2+9x-10$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{2x-5}{2-x}$
The domain of $\dfrac{f}{g}$ is the set of all real numbers except the zeros of $g$:
$2-x=0\Rightarrow x=2$
The domain is:
$(-\infty,2)\cup(2,\infty)$
