domain of sqrt(-x^3-x^2+16x+16) — step-by-step solution

1) Let's find the domain of

\sqrt{- x^{3} - x^{2} + 16 x + 16}

For the expression

\sqrt{- x^{3} - x^{2} + 16 x + 16}

to exist, you must check:

- x^{3}

- x^{2}

+ 16 x

+ 16 \geq 0

⇥1.1) Let's start with the obvious candidate:

x = -1

⇥Let's evaluate the value of:

- {(-1)}^{3}

- {(-1)}^{2}

+ 16 \cdot (-1)

+ 16

⇥ⓘ An integer power is a repeated product:

- {(-1)}^{3}

- ((-1) \cdot (-1) \cdot (-1))

- (-1)

■Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-1)}^{2}

(-1) \cdot (-1)

1

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

16 \cdot (-1)

-16

⇥

- (-1)

- 1

- 16

+ 16

⇥ⓘ Subtracting a negative number is adding it, two minus signs make a plus:

- (-1)

1

■Start from

-16

and move forward

16

on the number line, up to zero:

- 16 + 16

0

⇥

1 - 1 + 0

⇥ⓘ Adding 0 changes nothing:

- 1 + 0

-1

■Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

⇥

0

⇥Solution found:

x = -1

⇥1.2) Let's start the euclidean division of

(- x^{3}

- x^{2}

+ 16 x

+ 16)

(x + 1)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$- x^{3}$ $- x^{2}$ $+ 16 x$ $+ 16$	$- \frac{x^{3}}{x}$	$- x^{2}$	$- x^{2} \cdot (x + 1) = - x^{3} - x^{2}$	$(- x^{3}$ $- x^{2}$ $+ 16 x$ $+ 16) - (- x^{3} - x^{2}) = 16 x + 16$
$16 x + 16$	$\frac{16 x}{x}$	$16$	$16 \cdot (x + 1) = 16 x + 16$	$(16 x + 16) - (16 x + 16) = 0$

⇥Therefore, the result of the euclidean division is:

- x^{2} + 16

(x + 1) \cdot (16 - x^{2})

We study the sign of each factor of the product

⇥The expression

x + 1

is zero at

x = -1

, it is negative before and positive after

⇥1.3) Let's study the sign of

16 - x^{2}

by solving in

ℝ

16 - x^{2} \geq 0

⇥We divide both sides by

-1

; since this number is negative, we reverse the inequality sign

⇥An even power only depends on the absolute value:

x^{2} \leq 16

is equivalent to

| x | \leq 4

⇥

| A | \leq B

is equivalent to

- B \leq A \leq B

⇥

-4 \leq x \leq 4

⇥Solution obtained:

x \in [-4, 4]

⇥So the expression

16 - x^{2}

is positive or zero on

[-4, 4]

and negative otherwise

2) Let's draw up the sign table:

x	$-∞$		$-4$		$-1$		$4$		$+∞$
$x + 1$		-		-	0	+		+
$16 - x^{2}$		-	0	+		+	0	-
$- x^{3}$ $- x^{2}$ $+ 16 x$ $+ 16$		+	0	-	0	+	0	-

▶Domain of definition:

x \in (-∞, -4] \cup [-1, 4]