study (x^2+1)/(x-1)

Let's study the function

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

1) Let's evaluate the domain of

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

For the expression

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

to be defined, the following condition is necessary:

x - 1 \neq 0

ⓘ We add

1

to both sides

Solution obtained:

x \neq 1

Domain of definition:

x \neq 1

2) Let's study the parity of

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

with respect to

x

The domain is

x \neq 1

This domain is not symmetric about 0: the function is neither even nor odd

Parity: neither even nor odd

3) Differentiability

⇥For

p1

x^{2} + 1

and

q2

x - 1

, use the differentiation rule:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

⇥3.1) Let's perform the differentiation

p1

\frac{d}{d x} (x^{2} + 1)

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x^{2} + 1)

\frac{d}{d x} (x^{2}) + 0

⇥

\frac{d}{d x} (x^{2}) + 0

⇥ⓘ Adding 0 changes nothing:

\frac{d}{d x} (x^{2}) + 0

\frac{d}{d x} (x^{2})

■Differentiating

x^{n}

gives

n x^{n - 1}

\frac{d}{d x} (x^{2})

2 x

⇥

2 x

⇥3.2) Let's perform the differentiation

q2

\frac{d}{d x} (x - 1)

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x - 1)

1 + 0

⇥

1 + 0

⇥ⓘ Adding 0 changes nothing:

1 + 0

1

⇥

1

⇥3.3) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

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

⇥

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

⇥ⓘ With a minus sign before parentheses, change the sign of every term inside:

- (x^{2} + 1)

- x^{2} - 1

⇥

\frac{2 x \cdot (x - 1) - x^{2} - 1}{{(x - 1)}^{2}}

⇥ⓘ Expand using the distributive law, multiplying each term of the sum by the other factor:

2 x \cdot (x - 1)

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

⇥

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

⇥ⓘ The minus sign is placed in front of the product:

2 \cdot (- x)

- 2 x

■Group like terms by adding their coefficients:

2 x^{2} - x^{2}

x^{2}

⇥

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

⇥3.4) Then you can continue the differentiation:

ⅰ

\frac{d}{d x} (\frac{x^{2} + 1}{x - 1})

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

Derivative:

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

4) Let's study the variations of

f (x)

: first find where its derivative

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

is zero

We solve in

ℝ

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

⇥4.1) We solve in

ℝ

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

⇥Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -2

and

c = -1

⇥

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot (-1)

⇥ⓘ Calculate powers before multiplications, additions and subtractions. A power with a positive integer exponent is a repeated product:

{(-2)}^{2}

(-2) \cdot (-2)

4

⇥

Δ = 4 - 4 \cdot (-1)

⇥ⓘ Perform multiplications before additions and subtractions. In a product, two minus signs cancel: multiplying two negative numbers gives a positive number:

- 4 \cdot (-1)

4

⇥

Δ = 4 + 4

⇥ⓘ Start from

4

and move forward

4

on the number line:

4 + 4

8

⇥

Δ = 8

⇥Since

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-2) - \sqrt{8}}{2 \cdot 1}

⇥

x_{1} = \frac{2 - \sqrt{8}}{2}

⇥ⓘ We look for the largest square dividing

8

4

2^{2}

, and take it out of the root:

\sqrt{8}

\sqrt{4 \cdot 2}

\sqrt{4} \cdot \sqrt{2}

2 \cdot \sqrt{2}

⇥

x_{1} = \frac{2 - 2 \cdot \sqrt{2}}{2}

⇥ⓘ To simplify a fraction, divide its numerator and denominator by the same nonzero number:

\frac{2 - 2 \cdot \sqrt{2}}{2}

\frac{2 \cdot (1 - \sqrt{2})}{2}

1 - \sqrt{2}

⇥

x_{1} = 1 - \sqrt{2}

⇥The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-2) + \sqrt{8}}{2 \cdot 1}

⇥

x_{2} = \frac{2 + \sqrt{8}}{2}

⇥ⓘ We look for the largest square dividing

8

4

2^{2}

, and take it out of the root:

\sqrt{8}

\sqrt{4 \cdot 2}

\sqrt{4} \cdot \sqrt{2}

2 \cdot \sqrt{2}

⇥

x_{2} = \frac{2 + 2 \cdot \sqrt{2}}{2}

⇥ⓘ To simplify a fraction, divide its numerator and denominator by the same nonzero number:

\frac{2 + 2 \cdot \sqrt{2}}{2}

\frac{2 \cdot (1 + \sqrt{2})}{2}

1 + \sqrt{2}

⇥

x_{2} = 1 + \sqrt{2}

⇥Solutions obtained:

x = {1 - \sqrt{2}, 1 + \sqrt{2}}

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

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

⇥

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

means that:

x - 1 \neq 0

⇥ⓘ We add

1

to both sides

⇥Domain condition:

x \neq 1

f^{'} (x)

is zero at

x = {1 - \sqrt{2}, 1 + \sqrt{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^{'} (-1) = \frac{1}{2}

> 0:

f

is increasing on

(-∞, 1 - \sqrt{2})

f^{'} (0) = -1

< 0:

f

is decreasing on

(1 - \sqrt{2}, 1)

f^{'} (2) = -1

< 0:

f

is decreasing on

(1, 1 + \sqrt{2})

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

> 0:

f

is increasing on

(1 + \sqrt{2}, +∞)

Let's calculate the values of

f

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

When

x

tends to

-∞

f (x)

tends to

-∞

f (1 - \sqrt{2}) = \frac{{(1 - \sqrt{2})}^{2} + 1}{1 - \sqrt{2} - 1} = \sqrt{2} \cdot (\sqrt{2} - 2)

When

x

approaches

1

from the left,

f (x)

tends to

-∞

; from the right, to

+∞

f (1 + \sqrt{2}) = \frac{{(1 + \sqrt{2})}^{2} + 1}{1 + \sqrt{2} - 1} = \sqrt{2} \cdot (2 + \sqrt{2})

When

x

tends to

+∞

f (x)

tends to

+∞

Let's draw up the variation table:

$x$	$-∞$		$1 - \sqrt{2}$		$1$		$1 + \sqrt{2}$		$+∞$
$f^{'} (x)$		+	$0$	-	‖	-	$0$	+
$f (x)$	$-∞$	↗	$\sqrt{2} \cdot (\sqrt{2} - 2)$	↘	$-∞$ ‖ $+∞$	↘	$\sqrt{2} \cdot (2 + \sqrt{2})$	↗	$+∞$

Thus,

f

is increasing on

(-∞, 1 - \sqrt{2}]

, decreasing on

[1 - \sqrt{2}, 1)

and on

(1, 1 + \sqrt{2}]

and increasing on

[1 + \sqrt{2}, +∞)

5) Let's read the range from the variation table

⇥On

(-∞, 1)

f

is continuous: it takes every value of

(-∞, \sqrt{2} \cdot (\sqrt{2} - 2)]

⇥On

(1, +∞)

f

is continuous: it takes every value of

[\sqrt{2} \cdot (2 + \sqrt{2}), +∞)

Range:

(-∞, \sqrt{2} \cdot (\sqrt{2} - 2)] \cup [\sqrt{2} \cdot (2 + \sqrt{2}), +∞)

6) On the horizontal axis, y = 0: solve

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

We solve in

ℝ

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

⇥6.1) We solve in

ℝ

x^{2} + 1 = 0

⇥

x^{2} = -1

⇥In ℝ, a square is never negative: no real number has

-1

as its square

⇥Solution:

x \in \emptyset

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

x - 1 \neq 0

⇥Domain condition:

x \neq 1

The graph does not intersect the horizontal axis

On the vertical axis,

x = 0

: calculate f(0)

⇥

\frac{0^{2} + 1}{0 - 1}

⇥ⓘ Adding 0 changes nothing:

0 + 1

1

0 - 1

-1

■The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{1}{-1}

- \frac{1}{1}

⇥

- \frac{1}{1}

⇥

-1

Axis intercepts:

(0, -1)

7) Let's determine the asymptotes of

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

, for real

x

Vertical asymptotes: find the excluded values, where the denominator is zero. Then check whether the curve goes to infinity near these values

⇥7.1) We solve in

ℝ

x - 1 = 0

⇥ⓘ We add

1

to both sides

⇥Solution obtained:

x = 1

When

x

approaches

1

from the left (smaller values), the function tends to

-∞

; from the right (larger values), it tends to

+∞

The function tends to infinity on at least one side: the line

x = 1

is therefore a vertical asymptote

⇥7.2) Perform the Euclidean division of

(x^{2} + 1)

(x - 1)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{2} + 1$	$\frac{x^{2}}{x}$	$x$	$x \cdot (x - 1) = x^{2} - x$	$(x^{2} + 1) - (x^{2} - x) = x + 1$
$x + 1$	$\frac{x}{x}$	$1$	$x - 1 = x - 1$	$(x + 1) - (x - 1) = 2$

⇥Therefore, the result of the Euclidean division is:

(x + 1)

The division lets us write

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

: the line to test is

y = x + 1

For a slant asymptote, the difference between the function and the line must tend to 0: check the limit of

\frac{2}{x - 1}

at both infinities

The difference tends to

0

at +∞ and to

0

at -∞: the curve therefore approaches the line at both infinities

The slant asymptote is

y = x + 1

; therefore, there is no horizontal asymptote

🖼/draw?f1=%28x%5E2+%2B+1%29%2F%28x+-+1%29&v=x&x2=1&f3=x+%2B+1&b=-9.0&e=11.0&min=-10.0&max=14.0&dash=2,3

Asymptotes:

{x = 1, y = x + 1}

▶Study of $\frac{x^{2} + 1}{x - 1}$ : Domain of definition: $x \neq 1$ ; Parity: neither even nor odd; Derivative: $\frac{x^{2} - 2 x - 1}{{(x - 1)}^{2}}$ ; Range: $(-∞, \sqrt{2} \cdot (\sqrt{2} - 2)] \cup [\sqrt{2} \cdot (2 + \sqrt{2}), +∞)$ ; Axis intercepts: $(0, -1)$ ; Asymptotes: ${x = 1, y = x + 1}$

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