1/(sqrt(x^2-1))+3>= 0 — step-by-step solution

1) Let's solve in

ℝ

\frac{1}{\sqrt{x^{2} - 1}} + 3 \geq 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1}{\sqrt{x^{2} - 1}} + 3

\frac{1 + 3 \cdot \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}}

\frac{1 + 3 \cdot \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} \geq 0

We study the sign of the numerator

⇥1.1) Let's study the sign of

1 + 3 \cdot \sqrt{x^{2} - 1}

by solving in

ℝ

1 + 3 \cdot \sqrt{x^{2} - 1} \geq 0

⇥Isolate the square root:

\sqrt{x^{2} - 1} \geq - \frac{1}{3}

. Since a square root is always nonnegative,

\sqrt{x^{2} - 1}

is greater than

- \frac{1}{3}

wherever it is defined:

⇥The simplified relation is always true; the original expression's domain must still be respected

⇥So the expression

1 + 3 \cdot \sqrt{x^{2} - 1}

is always positive or zero

We study the sign of the denominator, which must not be zero

⇥1.2) Let's study the sign of

\sqrt{x^{2} - 1}

by solving in

ℝ

\sqrt{x^{2} - 1} > 0

⇥

\sqrt{x^{2} - 1} > 0

means that:

x^{2} - 1 > 0

⇥An even power only depends on the absolute value:

x^{2} > 1

is equivalent to

| x | > 1

⇥

| A | > B

is equivalent to:

A > B

A < - B

2) Let's draw up the sign table:

Candidates, pending the domain check:

x \in (-∞, -1) \cup (1, +∞)

3) Let's find the domain of

\frac{1}{\sqrt{x^{2} - 1}} + 3 \geq 0

For the expression

\frac{1}{\sqrt{x^{2} - 1}} + 3 \geq 0

to exist, you must check:

x^{2} - 1 > 0

An even power only depends on the absolute value:

x^{2} > 1

is equivalent to

| x | > 1

| A | > B

is equivalent to:

A > B

A < - B

▶Solution:

x \in (-∞, -1) \cup (1, +∞)