range of f(x)=1/(x^2-1) — step-by-step solution

1) Let's find the real values taken by

\frac{1}{x^{2} - 1}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = - \frac{2 x}{{(x^{2} - 1)}^{2}}

is zero

⇥Let's solve in

ℝ

- \frac{2 x}{{(x^{2} - 1)}^{2}} = 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) Let's solve in

ℝ

- 2 x = 0

⇥⇥We simplify the expression by

-2

⇥⇥⇥Let's solve in

ℝ

x = 0

⇥⇥⇥Candidates, pending the domain check:

x = 0

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

{(x^{2} - 1)}^{2} \neq 0

⇥⇥

{(x^{2} - 1)}^{2} \neq 0

is equivalent to:

x^{2} - 1 \neq 0

⇥⇥The two numbers whose square is

1

are

-1

and

1

⇥⇥Solutions obtained:

x \notin {-1, 1}

⇥ For the expression

- \frac{2 x}{{(x^{2} - 1)}^{2}}

to be defined, we must have:

{(x^{2} - 1)}^{2} \neq 0

⇥

{(x^{2} - 1)}^{2} \neq 0

has already been solved:

x \notin {-1, 1}

⇥Domain constraints:

x \in ℝ \ {-1, 1}

⇥

f^{'} (x)

is zero at

x = 0

⇥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^{'} (-2) = \frac{4}{9}

> 0:

f

is increasing on

(-∞, -1)

⇥

f^{'} (- \frac{1}{2}) = \frac{16}{9}

> 0:

f

is increasing on

(-1, 0)

⇥

f^{'} (\frac{1}{2}) = - \frac{16}{9}

< 0:

f

is decreasing on

(0, 1)

⇥

f^{'} (2) = - \frac{4}{9}

< 0:

f

is decreasing on

(1, +∞)

⇥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

0

⇥As

x

approaches

-1

from the left,

f (x)

tends to

+∞

; from the right, to

-∞

⇥

f (0) = \frac{1}{0^{2} - 1} = -1

⇥As

x

approaches

1

from the left,

f (x)

tends to

-∞

; from the right, to

+∞

⇥As

x

tends to

+∞

f (x)

tends to

0

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$-1$		$0$		$1$		$+∞$
$f^{'} (x)$		+	‖	+	$0$	-	‖	-
$f (x)$	$0$	↗	$+∞$ ‖ $-∞$	↗	$-1$	↘	$-∞$ ‖ $+∞$	↘	$0$

⇥So

f

is increasing on

(-∞, -1)

and on

(-1, 0]

and decreasing on

[0, 1)

and on

(1, +∞)

(-∞, -1)

f

is continuous: it takes every value of

(0, +∞)

(-1, 1)

f

is continuous: it takes every value of

(-∞, -1]

(1, +∞)

f

is continuous: it takes every value of

(0, +∞)

▶Range:

(-∞, -1] \cup (0, +∞)