extreme x-8/(x^2) — step-by-step solution

1) Find the strict local extrema of

x - \frac{8}{x^{2}}

: they are the points of the domain where its derivative changes sign

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = \frac{16}{x^{3}} + 1

is zero

⇥⇥1.1) Let's solve in

ℝ

\frac{16}{x^{3}} + 1 = 0

⇥⇥We multiply both sides by

x^{3}

, nonzero for the fractions to be defined

⇥⇥

\frac{16}{x^{3}} \cdot x^{3} + x^{3} = 0 \cdot x^{3}

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

\frac{16}{x^{3}} \cdot x^{3}

\frac{16 x^{3}}{x^{3}}

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

x^{3}

\frac{16 x^{3}}{x^{3}}

16

■A product with a zero factor is zero:

0 \cdot x^{3}

0

⇥⇥ⅰ

\frac{16}{x^{3}} + 1

16 + x^{3}

⇥⇥

16 + x^{3} = 0

⇥⇥Take the root:

- \sqrt[3]{16} = - 2 \cdot \sqrt[3]{2}

. The power being odd,

x^{3} = -16

has a single real solution,

- 2 \cdot \sqrt[3]{2}

⇥⇥Candidates, pending the domain check:

x = - 2 \cdot \sqrt[3]{2}

⇥⇥1.2) Let's find the domain of

\frac{16}{x^{3}} + 1 = 0

⇥⇥ For the expression

\frac{16}{x^{3}}

to be defined, we must have:

x^{3} \neq 0

⇥⇥

x^{3} \neq 0

is equivalent to:

x \neq 0

⇥⇥Domain constraints:

x \neq 0

⇥

f^{'} (x)

is zero at

x = - 2 \cdot \sqrt[3]{2}

⇥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^{'} (-3) = \frac{11}{27}

> 0:

f

is increasing on

(-∞, - 2 \cdot \sqrt[3]{2})

⇥

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

< 0:

f

is decreasing on

(- 2 \cdot \sqrt[3]{2}, 0)

⇥

f^{'} (1) = 17

> 0:

f

is increasing on

(0, +∞)

⇥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 (- 2 \cdot \sqrt[3]{2}) = - 2 \cdot \sqrt[3]{2} - \frac{8}{{(- 2 \cdot \sqrt[3]{2})}^{2}} = - 3 \cdot \sqrt[3]{2}

⇥As

x

approaches

0

from the left,

f (x)

tends to

-∞

; from the right, to

-∞

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$- 2 \cdot \sqrt[3]{2}$		$0$		$+∞$
$f^{'} (x)$		+	$0$	-	‖	+
$f (x)$	$-∞$	↗	$- 3 \cdot \sqrt[3]{2}$	↘	$-∞$ ‖ $-∞$	↗	$+∞$

⇥So

f

is increasing on

(-∞, - 2 \cdot \sqrt[3]{2}]

, decreasing on

[- 2 \cdot \sqrt[3]{2}, 0)

and increasing on

(0, +∞)

x = - 2 \cdot \sqrt[3]{2}

f

' changes from + to -:

f

has a strict local maximum,

f (- 2 \cdot \sqrt[3]{2}) = - 3 \cdot \sqrt[3]{2}

▶Strict local extrema:

{(- 2 \cdot \sqrt[3]{2}, - 3 \cdot \sqrt[3]{2})}