We want to with the functions $f(x)$ and $g(x).$

### Calculating $f(x)+g(x)$

Let's substitute the given function rules into the expressions

$f(x)$ and

$g(x).$ Then we will perform the addition.

$f(x)+g(x)$

$(x−1)+(5x−2)$

$x−1+5x−2$

$6x−3$

After we have performed the addition, the result takes the form of

$6x−3.$
$f(x)+g(x)=6x−3 $ ### Calculating $f(x)−g(x)$

Let's substitute the given function rules into the expressions

$f(x)$ and

$g(x).$ Then we will perform the subtraction.

$f(x)−g(x)$

$(x−1)−(5x−2)$

$x−1−5x+2$

$-4x+1$

After we have performed the subtraction, the result takes the form of

$-4x+1.$
$f(x)−g(x)=-4x+1 $ ### Calculating $f(x)⋅g(x)$

Let's substitute the given function rules into the expressions

$f(x)$ and

$g(x).$ Then we will perform the multiplication.

$f(x)⋅g(x)$

$(x−1)⋅(5x−2)$

$x(5x−2)−1(5x−2)$

$5x_{2}−2x−5x+2$

$5x_{2}−7x+2$

After we have performed the multiplication, the result takes the form of

$5x_{2}−7x+2.$
$f(x)⋅g(x)=5x_{2}−7x+2 $ ### Calculating $g(x)f(x) $

To perform the division, let's substitute the given function rules into the expressions

$f(x)$ and

$g(x).$
This expression cannot be simplified any further. The of both

$f(x)$ and

$g(x)$ is all ,

$R.$ The domain of

$g(x)f(x) $ is the set of numbers common to the domains of both

$f(x)$ and

$g(x).$ However, the denominator

**cannot** be

$0.$ Let's find the values that make

$g(x)=0.$
$g(x)=0$

$5x−2=0$

$5x=2$

$x=52 $

As we can see,

$x=52 $ makes

$g(x)=0.$ Therefore, this value must be excluded from the domain of the quotient. We can say that the domain is the set of all real numbers except

$52 .$
$g(x)f(x) =5x−2x−1 ,x =52 $