domain of f(x)=(x^3-x)/(x^3-x^2-2x)

We save the definition:

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

1) Let's evaluate the domain of

\frac{x^{3} - x}{x^{3} - x^{2} - 2 x}

For the expression

\frac{x^{3} - x}{x^{3} - x^{2} - 2 x}

to be defined, we must have:

x^{3} - x^{2} - 2 x \neq 0

You can factorize the expression:

x^{3} - x^{2} - 2 x

into:

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

⇥1.1) Let's solve in

ℝ

x \neq 0

⇥Solution found:

x \neq 0

⇥1.2) Let's solve in

ℝ

x^{2} - x - 2 \neq 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 = -1

and

c = -2

⇥

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-2)

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

{(-1)}^{2}

(-1) \cdot (-1)

1

⇥

Δ = 1 - 4 \cdot (-2)

⇥ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-2)

8

⇥

Δ = 1 + 8

⇥ⓘ Start from

1

and move forward

8

on the number line:

1 + 8

9

⇥

Δ = 9

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{9}}{2 \cdot 1}

⇥

x_{1} = \frac{1 - \sqrt{9}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

⇥

x_{1} = \frac{1 - 3}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

1 - 3

-2

⇥

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

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{2}

-1

⇥

x_{1} = -1