extreme f(x)=(x^2+x+1)/x — step-by-step solution

1) Find the strict local extrema of

\frac{x^{2} + x + 1}{x}

: 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{x^{2} - 1}{x^{2}}

is zero

⇥Let's solve in

ℝ

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

⇥A quotient is zero if and only if its numerator is zero and its denominator is not: we solve the numerator equal to 0

⇥⇥1.1) Let's solve in

ℝ

x^{2} - 1 = 0

⇥⇥The two numbers whose square is

1

are

-1

and

1

⇥⇥Candidates, pending the domain check:

x = {-1, 1}

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

x^{2} \neq 0

⇥⇥

x^{2} \neq 0

means that:

x \neq 0

⇥⇥Domain condition:

x \neq 0

⇥ For the expression

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

to be defined, we must have:

x^{2} \neq 0

⇥Domain constraints:

x \neq 0

⇥

f^{'} (x)

is zero at

x = {-1, 1}

⇥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^{'} (-2) = \frac{3}{4}

> 0:

f

is increasing on

(-∞, -1)

⇥

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

< 0:

f

is decreasing on

(-1, 0)

⇥

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

< 0:

f

is decreasing on

(0, 1)

⇥

f^{'} (2) = \frac{3}{4}

> 0:

f

is increasing on

(1, +∞)

⇥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 (-1) = \frac{{(-1)}^{2} - 1 + 1}{-1} = -1

⇥As

x

approaches

0

from the left,

f (x)

tends to

-∞

; from the right, to

+∞

⇥

f (1) = \frac{1^{2} + 1 + 1}{1} = 3

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$-1$		$0$		$1$		$+∞$
$f^{'} (x)$		+	$0$	-	‖	-	$0$	+
$f (x)$	$-∞$	↗	$-1$	↘	$-∞$ ‖ $+∞$	↘	$3$	↗	$+∞$

⇥So

f

is increasing on

(-∞, -1]

, decreasing on

[-1, 0)

and on

(0, 1]

and increasing on

[1, +∞)

x = -1

f

' changes from + to -:

f

has a strict local maximum,

f (-1) = -1

x = 1

f

' changes from - to +:

f

has a strict local minimum,

f (1) = 3

▶Strict local extrema:

{(-1, -1), (1, 3)}