domain of f(x)=(sqrt(2-x))/(x^2+4x)

We save the definition:

f (x) = \frac{\sqrt{2 - x}}{x^{2} + 4 x}

1) Let's evaluate the domain of

\frac{\sqrt{2 - x}}{x^{2} + 4 x}

For the expression

\frac{\sqrt{2 - x}}{x^{2} + 4 x}

to exist, you must check:

x^{2} + 4 x \neq 0

For the expression

\sqrt{2 - x}

to exist, you must check:

2 - x \geq 0

2) Let's solve in

ℝ

x^{2} + 4 x \neq 0

You can factorize the expression:

x^{2} + 4 x

into:

x \cdot (x + 4)

⇥2.1) Let's solve in

ℝ

x \neq 0

⇥Solution found:

x \neq 0

⇥2.2) Let's solve in

ℝ

x + 4 \neq 0

⇥Solution obtained:

x \neq -4

3) Let's solve in

ℝ

2 - x \geq 0

ⓘ We add

x

to both sides

2 \geq x

Solution obtained:

x \leq 2

Let's create the table of the different solutions to find the overall solution:

▶Domain of definition:

x \in (-∞, -4) \cup (-4, 0) \cup (0, 2]