critical points of x^3-6x^2+9x+1

1) A polynomial is differentiable on ℝ: its critical points are the zeros of its derivative

⇥1.1) Let's perform the differentiation:

\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x + 1)

⇥ⓘ Differentiate a sum by differentiating each term, then adding the derivatives:

\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x + 1)

\frac{d}{d x} (x^{3})

+ \frac{d}{d x} (- 6 x^{2})

+ \frac{d}{d x} (9 x)

+ 0

⇥

\frac{d}{d x} (x^{3})

+ \frac{d}{d x} (- 6 x^{2})

+ \frac{d}{d x} (9 x)

+ 0

⇥ⓘ Differentiating

x^{n}

gives

n x^{n - 1}

\frac{d}{d x} (x^{3})

3 x^{2}

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

2 x

■The derivative of a constant multiple keeps that factor: that of

k u

k \cdot u'

\frac{d}{d x} (- 6 x^{2})

-6 \cdot \frac{d}{d x} (x^{2})

\frac{d}{d x} (9 x)

9

■Adding 0 changes nothing:

\frac{d}{d x} (9 x) + 0

\frac{d}{d x} (9 x)

⇥

3 x^{2} - 6 \cdot 2 x + 9

⇥ⓘ Perform multiplications before additions and subtractions:

6 \cdot 2

12

⇥

3 x^{2} - 12 x + 9

⇥1.2) Let's solve in

ℝ

3 x^{2} - 12 x + 9 = 0

⇥Since

3 \neq 0

, divide both sides by

3

x^{2} - 4 x + 3 = 0

⇥Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -4

and

c = 3

⇥

Δ = {(-4)}^{2} - 4 \cdot 1 \cdot 3

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

{(-4)}^{2}

(-4) \cdot (-4)

16

⇥

Δ = 16 - 4 \cdot 3

⇥ⓘ Perform multiplications before additions and subtractions:

4 \cdot 3

12

⇥

Δ = 16 - 12

⇥ⓘ Start from

16

and move back

12

on the number line:

16 - 12

4

⇥

Δ = 4

⇥Since

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-4) - \sqrt{4}}{2 \cdot 1}

⇥

x_{1} = \frac{4 - \sqrt{4}}{2}

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

\sqrt{4}

2

because

2^{2}

4

⇥

x_{1} = \frac{4 - 2}{2}

⇥ⓘ Calculate the numerator and denominator before dividing:

4 - 2

2

⇥

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

⇥ⓘ Simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

⇥

x_{1} = 1

⇥The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-4) + \sqrt{4}}{2 \cdot 1}

⇥

x_{2} = \frac{4 + \sqrt{4}}{2}

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

\sqrt{4}

2

because

2^{2}

4

⇥

x_{2} = \frac{4 + 2}{2}

⇥ⓘ Calculate the numerator and denominator before dividing:

4 + 2

6

⇥

x_{2} = \frac{6}{2}

⇥ⓘ Simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

⇥

x_{2} = 3

▶Solutions obtained:

x = {1, 3}

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