range of 7+sqrt(6+x) — step-by-step solution

1) Let's find the real values taken by

7 + \sqrt{6 + x}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = \frac{1}{2 \cdot \sqrt{x + 6}}

is zero

⇥⇥1.1) Let's solve in

ℝ

\frac{1}{2 \cdot \sqrt{x + 6}} = 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 \sqrt{x + 6} \neq 0

⇥⇥⇥We simplify the expression by

2

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

\sqrt{x + 6} \neq 0

⇥⇥⇥

\sqrt{x + 6} \neq 0

is equivalent to:

x + 6 \neq 0

⇥⇥⇥ⓘ We subtract

6

from both sides

⇥⇥⇥Domain condition:

x \neq -6

⇥⇥1.2) Let's evaluate the domain of

\frac{1}{2 \cdot \sqrt{x + 6}} = 0

⇥⇥ For the expression

\frac{1}{2 \cdot \sqrt{x + 6}}

to be defined, we must have:

2 \cdot \sqrt{x + 6} \neq 0

⇥⇥ For the expression

\sqrt{x + 6}

to be defined, we must have:

x + 6 \geq 0

⇥⇥

2 \cdot \sqrt{x + 6} \neq 0

has already been solved:

x \neq -6

⇥⇥1.3) Let's solve in

ℝ

x + 6 \geq 0

⇥⇥ⓘ We subtract

6

from both sides

⇥⇥Solution found:

x \geq -6

⇥⇥Intersecting

x > -6

and

x \geq -6

gives

x > -6

⇥⇥Domain constraints:

x > -6

⇥

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{6}}{12}

> 0:

f

is increasing on

(-6, +∞)

⇥Let's calculate the values of

f

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

⇥

f (-6) = 7 + \sqrt{6 - 6} = 7

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

$x$	$-6$		$+∞$
$f^{'} (x)$	‖	+
$f (x)$	$7$	↗	$+∞$

⇥So

f

is increasing on

[-6, +∞)

[-6, +∞)

f

is continuous: it takes every value of

[7, +∞)

▶Range:

[7, +∞)