range of f(x)=4/(x-5) — step-by-step solution

1) Let's find the real values taken by

\frac{4}{x - 5}

, for real

x

⇥Let's find the domain of

\frac{4}{x - 5}

⇥ For the expression

\frac{4}{x - 5}

to exist, you must check:

x - 5 \neq 0

⇥ⓘ We add

5

to both sides

⇥Solution obtained:

x \neq 5

The input domain is

x \neq 5

2) Set

y = \frac{4}{x - 5}

and solve for

x

to express the input in terms of the attained value

⇥2.1) Let's solve in

ℝ

y = \frac{4}{x - 5}

⇥

\frac{4}{x - 5} = y

⇥We multiply both sides by

x - 5

, nonzero for the fractions to be defined

⇥

\frac{4}{x - 5} \cdot (x - 5) = y \cdot (x - 5)

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{4}{x - 5} \cdot (x - 5)

\frac{4 \cdot (x - 5)}{x - 5}

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor

x - 5

\frac{4 \cdot (x - 5)}{x - 5}

4

⇥

4 = y \cdot (x - 5)

⇥ⓘ Expand by multiplying each term inside the parentheses:

y \cdot (x - 5)

y x + y \cdot (-5)

⇥

4 = y x + y \cdot (-5)

⇥ⓘ The minus sign is placed in front of the product:

y \cdot (-5)

- 5 y

⇥

4 = y x - 5 y

⇥

y x - 5 y = 4

⇥

y x = 4 + 5 y

⇥ⓘ Both sides are divided by

y

, which must be nonzero:

y \neq 0

⇥Candidates, pending the domain check:

x = \frac{5 y + 4}{y}

⇥2.2) Let's evaluate the domain of

y = \frac{4}{x - 5}

⇥ For the expression

\frac{4}{x - 5}

to exist, you must check:

x - 5 \neq 0

⇥ⓘ We add

5

to both sides

⇥Domain constraints:

x \neq 5

3) The equation has a solution for

x

exactly when

y \neq 0

: these are the attained values

▶Range:

ℝ \ {0}