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

1) Let's find the real values taken by

\frac{1}{\sqrt{x + 2}}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = - \frac{1}{2 \cdot {(x + 2)}^{\frac{3}{2}}}

is zero

⇥⇥1.1) Let's solve in

ℝ

- \frac{1}{2 \cdot {(x + 2)}^{\frac{3}{2}}} = 0

⇥⇥⇥1.1.1) Let's solve in

ℝ

-1 = 0

⇥⇥⇥The statement

-1 = 0

is false, whatever the value of

x

⇥⇥⇥Solution:

x \in \emptyset

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

2 \cdot {(x + 2)}^{\frac{3}{2}} \neq 0

⇥⇥⇥We simplify the expression by

2

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

{(x + 2)}^{\frac{3}{2}} \neq 0

⇥⇥⇥

{(x + 2)}^{\frac{3}{2}} \neq 0

is equivalent to:

x + 2 \neq 0

⇥⇥⇥ⓘ We subtract

2

from both sides

⇥⇥⇥Domain condition:

x \neq -2

⇥⇥1.2) Let's find the domain of

- \frac{1}{2 \cdot {(x + 2)}^{\frac{3}{2}}} = 0

⇥⇥ For the expression

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

to exist, you must check:

2 \cdot {(x + 2)}^{\frac{3}{2}} \neq 0

⇥⇥ For the expression

{(x + 2)}^{\frac{3}{2}}

to be defined, we must have:

x + 2 \geq 0

⇥⇥

2 \cdot {(x + 2)}^{\frac{3}{2}} \neq 0

has already been solved:

x \neq -2

⇥⇥1.3) Let's solve in

ℝ

x + 2 \geq 0

⇥⇥ⓘ We subtract

2

from both sides

⇥⇥Solution obtained:

x \geq -2

⇥⇥Intersecting

x > -2

and

x \geq -2

gives

x > -2

⇥⇥Domain constraints:

x > -2

⇥

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^{'} (0) = - \frac{\sqrt{2}}{8}

< 0:

f

is decreasing 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

approaches

-2

from the right,

f (x)

tends to

+∞

⇥As

x

tends to

+∞

f (x)

tends to

0

⇥Let's draw up the variation table:

⇥

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

⇥So

f

is decreasing on

(-2, +∞)

(-2, +∞)

f

is continuous: it takes every value of

(0, +∞)

▶Range:

(0, +∞)