range of x+16/x — step-by-step solution

1) Let's find the real values taken by

x + \frac{16}{x}

using its variation table

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = 1 - \frac{16}{x^{2}}

is zero

⇥⇥1.1) Let's solve in

ℝ

1 - \frac{16}{x^{2}} = 0

⇥⇥We multiply both sides by

x^{2}

, nonzero for the fractions to be defined

⇥⇥

x^{2} - \frac{16}{x^{2}} \cdot x^{2} = 0 \cdot x^{2}

⇥⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{16}{x^{2}} \cdot x^{2}

\frac{16 x^{2}}{x^{2}}

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

- \frac{16 x^{2}}{x^{2}}

-16

■A product with a zero factor is zero:

0 \cdot x^{2}

0

⇥⇥ⅰ

1 - \frac{16}{x^{2}}

x^{2} - 16

⇥⇥

x^{2} - 16 = 0

⇥⇥The two numbers whose square is

16

are

-4

and

4

⇥⇥Candidates, pending the domain check:

x = {-4, 4}

⇥⇥1.2) Let's evaluate the domain of

1 - \frac{16}{x^{2}} = 0

⇥⇥ For the expression

- \frac{16}{x^{2}}

to exist, you must check:

x^{2} \neq 0

⇥⇥

x^{2} \neq 0

is equivalent to:

x \neq 0

⇥⇥Domain constraints:

x \neq 0

⇥

f^{'} (x)

is zero at

x = {-4, 4}

⇥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^{'} (-5) = \frac{9}{25}

> 0:

f

is increasing on

(-∞, -4)

⇥

f^{'} (-1) = -15

< 0:

f

is decreasing on

(-4, 0)

⇥

f^{'} (1) = -15

< 0:

f

is decreasing on

(0, 4)

⇥

f^{'} (5) = \frac{9}{25}

> 0:

f

is increasing on

(4, +∞)

⇥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 (-4) = - 4 + \frac{16}{-4} = -8

⇥As

x

approaches

0

from the left,

f (x)

tends to

-∞

; from the right, to

+∞

⇥

f (4) = 4 + \frac{16}{4} = 8

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$-4$		$0$		$4$		$+∞$
$f^{'} (x)$		+	$0$	-	‖	-	$0$	+
$f (x)$	$-∞$	↗	$-8$	↘	$-∞$ ‖ $+∞$	↘	$8$	↗	$+∞$

⇥So

f

is increasing on

(-∞, -4]

, decreasing on

[-4, 0)

and on

(0, 4]

and increasing on

[4, +∞)

(-∞, 0)

f

is continuous: it takes every value of

(-∞, -8]

(0, +∞)

f

is continuous: it takes every value of

[8, +∞)

▶Range:

(-∞, -8] \cup [8, +∞)