range of sqrt(2-(x-1)) — step-by-step solution

1) Let's find the real values taken by

\sqrt{2 - (x - 1)}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

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

is zero

⇥⇥1.1) Let's solve in

ℝ

- \frac{1}{2 \cdot \sqrt{3 - x}} = 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{3 - x} \neq 0

⇥⇥⇥We simplify the expression by

2

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

\sqrt{3 - x} \neq 0

⇥⇥⇥

\sqrt{3 - x} \neq 0

means that:

3 - x \neq 0

⇥⇥⇥ⓘ We add

x

to both sides

⇥⇥⇥

3 \neq x

⇥⇥⇥Domain condition:

x \neq 3

⇥⇥1.2) Let's find the domain of

- \frac{1}{2 \cdot \sqrt{3 - x}} = 0

⇥⇥ For the expression

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

to exist, you must check:

2 \cdot \sqrt{3 - x} \neq 0

⇥⇥ For the expression

\sqrt{3 - x}

to be defined, we must have:

3 - x \geq 0

⇥⇥

2 \cdot \sqrt{3 - x} \neq 0

has already been solved:

x \neq 3

⇥⇥1.3) Let's solve in

ℝ

3 - x \geq 0

⇥⇥ⓘ We add

x

to both sides

⇥⇥

3 \geq x

⇥⇥Solution found:

x \leq 3

⇥⇥Intersecting

x < 3

and

x \leq 3

gives

x < 3

⇥⇥Domain constraints:

x < 3

⇥

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{3}}{6}

< 0:

f

is decreasing on

(-∞, 3)

⇥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 (3) = \sqrt{2 - (3 - 1)} = 0

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$3$
$f^{'} (x)$		-	‖
$f (x)$	$+∞$	↘	$0$

⇥So

f

is decreasing on

(-∞, 3]

(-∞, 3]

f

is continuous: it takes every value of

[0, +∞)

▶Range:

[0, +∞)