range of pi-3arcsin(2x-1) — step-by-step solution

1) Let's find the real values taken by

π - 3 \cdot \arcsin (2 x - 1)

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

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

is zero

⇥⇥1.1) Let's solve in

ℝ

- \frac{6}{\sqrt{1 - {(2 x - 1)}^{2}}} = 0

⇥⇥⇥1.1.1) Let's solve in

ℝ

-6 = 0

⇥⇥⇥The statement

-6 = 0

is false, whatever the value of

x

⇥⇥⇥Solution:

x \in \emptyset

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

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

⇥⇥⇥

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

is equivalent to:

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

⇥⇥⇥For

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

, we recognize a difference of two squares:

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

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 1

and

b = 2 x - 1

⇥⇥⇥You can factor:

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

into:

(1 - (2 x - 1)) \cdot (1 + (2 x - 1))

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

(1 - (2 x - 1)) \cdot (1 + (2 x - 1)) \neq 0

⇥⇥⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (2 x - 1)

- 2 x + 1

■Parentheses are computed first:

1 - 1

0

⇥⇥⇥

(1 - 2 x + 1) \cdot (0 + 2 x) \neq 0

⇥⇥⇥ⓘ Parentheses are computed first:

1 + 1

2

■Adding 0 changes nothing:

0 + 2 x

2 x

⇥⇥⇥

(2 - 2 x) \cdot 2 x \neq 0

⇥⇥⇥We simplify the expression by

2

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

2 - 2 x \neq 0

⇥⇥⇥

2 x \neq 2

⇥⇥⇥ⓘ Both sides are divided by

2

⇥⇥⇥

x \neq \frac{2}{2}

⇥⇥⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

⇥⇥⇥Domain condition:

x \neq 1

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

x \neq 0

⇥⇥⇥Domain condition:

x \neq 0

⇥⇥1.2) Let's find the domain of

- \frac{6}{\sqrt{1 - {(2 x - 1)}^{2}}} = 0

⇥⇥ For the expression

- \frac{6}{\sqrt{1 - {(2 x - 1)}^{2}}} = 0

to be defined, we must have:

1 - {(2 x - 1)}^{2} > 0

⇥⇥ For the expression

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

to exist, you must check:

1 - (2 x - 1) \cdot (2 x - 1) \geq 0

⇥⇥1.3) Let's solve in

ℝ

1 - {(2 x - 1)}^{2} > 0

⇥⇥For

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

, we recognize a difference of two squares:

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

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 1

and

b = 2 x - 1

⇥⇥You can factor:

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

into:

(1 - (2 x - 1)) \cdot (1 + (2 x - 1))

⇥⇥1.4) Let's solve in

ℝ

(1 - (2 x - 1)) \cdot (1 + (2 x - 1)) > 0

⇥⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (2 x - 1)

- 2 x + 1

■Parentheses are computed first:

1 - 1

0

⇥⇥

(1 - 2 x + 1) \cdot (0 + 2 x) > 0

⇥⇥ⓘ Parentheses are computed first:

1 + 1

2

■Adding 0 changes nothing:

0 + 2 x

2 x

⇥⇥

(2 - 2 x) \cdot 2 x > 0

⇥⇥We simplify the expression by

2

⇥⇥We study the sign of each factor of the product

⇥⇥⇥The expression

2 - 2 x

is zero at:

⇥⇥⇥

x = \frac{-2}{-2}

⇥⇥⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-2}{-2}

\frac{2}{2}

⇥⇥⇥

x = \frac{2}{2}

⇥⇥⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

⇥⇥⇥

x = 1

⇥⇥⇥Its coefficient of

x

-2

, being negative, it is positive before and negative after

1

⇥⇥⇥The expression

x

is zero at

x = 0

, it is negative before and positive after

⇥⇥1.5) Let's draw up the sign table:

⇥⇥

x	$-∞$		$0$		$1$		$+∞$
$2 - 2 x$		+		+	0	-
$x$		-	0	+		+
$1 - {(2 x - 1)}^{2}$		-	0	+	0	-

⇥⇥Solution found:

x \in (0, 1)

⇥⇥1.6) Let's solve in

ℝ

1 - (2 x - 1) \cdot (2 x - 1) \geq 0

⇥⇥ⓘ Expand by multiplying each term inside the parentheses:

- (2 x - 1) \cdot (2 x - 1)

- (2 \cdot x \cdot 2 \cdot x

+ 2 x \cdot (-1)

- 1 \cdot 2 x

- 1 \cdot (-1))

⇥⇥

1 - (2 \cdot x \cdot 2 \cdot x

+ 2 x \cdot (-1)

- 1 \cdot 2 x

- 1 \cdot (-1)) \geq 0

⇥⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (2 \cdot x \cdot 2 \cdot x

+ 2 x \cdot (-1)

- 1 \cdot 2 x

- 1 \cdot (-1))

- 2 \cdot x \cdot 2 \cdot x

- 2 x \cdot (-1)

+ 1 \cdot 2 x

+ 1 \cdot (-1)

⇥⇥

1

- 2 \cdot x \cdot 2 \cdot x

- 2 x \cdot (-1)

+ 1 \cdot 2 x

+ 1 \cdot (-1) \geq 0

⇥⇥ⓘ Multiplications are done before additions and subtractions:

2 \cdot 2

4

■Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-2 \cdot (-1)

2

⇥⇥

1

- 4 x^{2}

+ 2 x

+ 2 x

- 1 \geq 0

⇥⇥ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

■Like terms are grouped by adding their coefficients:

2 x + 2 x

4 x

⇥⇥

0 - 4 x^{2} + 4 x \geq 0

⇥⇥ⓘ Adding 0 changes nothing:

0 - 4 x^{2}

-4 x^{2}

⇥⇥

- 4 x^{2} + 4 x \geq 0

⇥⇥You can factor:

4 x - 4 x^{2}

into:

x \cdot (4 - 4 x)

⇥⇥We study the sign of each factor of the product

⇥⇥⇥The expression

x

is zero at

x = 0

, it is negative before and positive after

⇥⇥⇥The expression

4 - 4 x

is zero at:

⇥⇥⇥

x = \frac{-4}{-4}

⇥⇥⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-4}{-4}

\frac{4}{4}

⇥⇥⇥

x = \frac{4}{4}

⇥⇥⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

4

, their greatest common divisor:

\frac{4}{4}

1

⇥⇥⇥

x = 1

⇥⇥⇥Its coefficient of

x

-4

, being negative, it is positive before and negative after

1

⇥⇥1.7) Let's draw up the sign table:

⇥⇥

x	$-∞$		$0$		$1$		$+∞$
$x$		-	0	+		+
$4 - 4 x$		+		+	0	-
$1 - (2 x - 1) \cdot (2 x - 1)$		-	0	+	0	-

⇥⇥Solution obtained:

x \in [0, 1]

⇥⇥Intersecting

x \in (0, 1)

and

x \in [0, 1]

gives

x \in (0, 1)

⇥⇥Domain constraints:

x \in (0, 1)

⇥

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^{'} (\frac{1}{2}) = -6

< 0:

f

is decreasing on

(0, 1)

⇥Let's calculate the values of

f

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

⇥

f (0) = π - 3 \cdot \arcsin (2 \cdot 0 - 1) = \frac{5 π}{2}

⇥

f (1) = π - 3 \cdot \arcsin (2 \cdot 1 - 1) = - \frac{π}{2}

⇥Let's draw up the variation table:

⇥

$x$	$0$		$1$
$f^{'} (x)$	‖	-	‖
$f (x)$	$\frac{5 π}{2}$	↘	$- \frac{π}{2}$

⇥So

f

is decreasing on

[0, 1]

[0, 1]

f

is continuous: it takes every value of

[- \frac{π}{2}, \frac{5 π}{2}]

▶Range:

[- \frac{π}{2}, \frac{5 π}{2}]