range of sqrt(x^2-4) — step-by-step solution

1) Let's find the real values taken by

\sqrt{x^{2} - 4}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

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

is zero

⇥⇥1.1) Let's solve in

ℝ

\frac{x}{\sqrt{x^{2} - 4}} = 0

⇥⇥A quotient is zero if and only if its numerator is zero and its denominator is not: we solve the numerator equal to 0

⇥⇥⇥1.1.1) Let's solve in

ℝ

x = 0

⇥⇥⇥Candidates, pending the domain check:

x = 0

⇥⇥⇥1.1.2) To prevent the denominator from being zero, let's solve:

\sqrt{x^{2} - 4} \neq 0

⇥⇥⇥

\sqrt{x^{2} - 4} \neq 0

is equivalent to:

x^{2} - 4 \neq 0

⇥⇥⇥The two numbers whose square is

4

are

-2

and

2

⇥⇥⇥Solutions found:

x \notin {-2, 2}

⇥⇥1.2) Let's find the domain of

\frac{x}{\sqrt{x^{2} - 4}} = 0

⇥⇥ For the expression

\frac{x}{\sqrt{x^{2} - 4}} = 0

to be defined, we must have:

x^{2} - 4 > 0

⇥⇥An even power only depends on the absolute value:

x^{2} > 4

is equivalent to

| x | > 2

⇥⇥

| A | > B

means that:

A > B

A < - B

⇥⇥Domain constraints:

x \in (-∞, -2) \cup (2, +∞)

⇥⇥The solution

x = 0

does not satisfy the domain

x \in (-∞, -2) \cup (2, +∞)

and is discarded

⇥

f^{'} (x)

is never zero on the domain of

f

⇥Built from usual functions,

f^{'} (x)

is continuous where it is defined: on each interval where it is never zero, it keeps the sign of its value at any point

⇥

f^{'} (-3) = - \frac{3 \cdot \sqrt{5}}{5}

< 0:

f

is decreasing on

(-∞, -2)

⇥

f^{'} (3) = \frac{3 \cdot \sqrt{5}}{5}

> 0:

f

is increasing on

(2, +∞)

⇥Let's calculate the values of

f

at the bounds of the table, or its limits at the infinite or excluded bounds

⇥As

x

tends to

-∞

f (x)

tends to

+∞

⇥

f (-2) = \sqrt{{(-2)}^{2} - 4} = 0

⇥

f (2) = \sqrt{2^{2} - 4} = 0

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$-2$	$2$		$+∞$
$f^{'} (x)$		-	‖	‖	+
$f (x)$	$+∞$	↘	$0$	$0$	↗	$+∞$

⇥So

f

is decreasing on

(-∞, -2]

and increasing on

[2, +∞)

[2, +∞)

f

is continuous: it takes every value of

[0, +∞)

▶Range:

[0, +∞)