range of f(x)= 1/(x+3) — step-by-step solution

1) Let's find the real values taken by

\frac{1}{x + 3}

, for real

x

⇥Let's evaluate the domain of

\frac{1}{x + 3}

⇥ For the expression

\frac{1}{x + 3}

to be defined, we must have:

x + 3 \neq 0

⇥Solution found:

x \neq -3

The input domain is

x \in ℝ \ {-3}

2) Set

y = \frac{1}{x + 3}

and solve for

x

to express the input in terms of the attained value

⇥2.1) Let's solve in

ℝ

\frac{1}{x + 3} = y

⇥Let's add fractions:

\frac{1}{x + 3} - y = 0

⇥

\frac{1 - y \cdot (x + 3)}{x + 3} = 0

⇥2.2) Let's solve in

ℝ

1 - y \cdot (x + 3) = 0

⇥

1 - (y x + y \cdot 3) = 0

⇥

1 - y \cdot x - y \cdot 3 = 0

⇥

- y x = - 1 + y \cdot 3

⇥

x = \frac{- 1 + y \cdot 3}{- y}

⇥

x = - \frac{- 1 + y \cdot 3}{y}

⇥

x = \frac{1 - y \cdot 3}{y}

⇥Solution found:

x = \frac{1 - 3 y}{y}

⇥2.3) To avoid a zero denominator, let's solve:

x + 3 \neq 0

⇥Solution excluded:

x \neq -3

3) Every value in the inverse's domain is attained, so its domain gives the range

⇥Let's find the domain of

\frac{1 - 3 y}{y}

⇥ For the expression

\frac{1 - 3 y}{y}

to be defined, we must have:

y \neq 0

⇥Solution found:

y \in ℝ \ {0}

▶Range:

ℝ \ {0}