F(x) = x + 3, for x ≠ 1.

Step-by-step explanation:

solve

3

−

x

=

0

⇒

x

=

3

←

excluded value

⇒

domain

x

∈

R

,

x

≠

3

x

∈

(

−

∞

,

3

)

∪

(

3

,

∞

)

←

in interval notation

f

(

x

)

=

3

+

x

3

−

x

divide terms on numerator/denominator by

x

f

(

x

)

=

3

x

+

x

x

3

x

−

x

x

=

3

x

+

1

3

x

−

1

as

x

→

±

∞

,

f

(

x

)

→

0

+

1

0

−

1

=

−

1

←

excluded value

⇒

range

y

∈

R

,

y

≠

−

1

y

∈

(

−

∞

,

−

1

)

∪

(

−

1

,

∞

)

graph{(3+x)/(3-x) [-10, 10, -5, 5]}solve

3

−

x

=

0

⇒

x

=

3

←

excluded value

⇒

domain

x

∈

R

,

x

≠

3

x

∈

(

−

∞

,

3

)

∪

(

3

,

∞

)

←

in interval notation

f

(

x

)

=

3

+

x

3

−

x

divide terms on numerator/denominator by

x

f

(

x

)

=

3

x

+

x

x

3

x

−

x

x

=

3

x

+

1

3

x

−

1

as

x

→

±

∞

,

f

(

x

)

→

0

+

1

0

−

1

=

−

1

←

excluded value

⇒

range

y

∈

R

,

y

≠

−

1

y

∈

(

−

∞

,

−

1

)

∪

(

−

1

,

∞

)

graph{(3+x)/(3-x) [-10, 10, -5, 5]}

step-by-step explanation: