extreme f(x)=x^3-12x+2 — step-by-step solution

1) Find the strict local extrema of

x^{3} - 12 x + 2

on ℝ: the derivative must be zero and change sign

⇥1.1) Let's differentiate:

\frac{d}{d x} (x^{3} - 12 x + 2)

⇥

\frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 12 x) + 0

⇥ℹ

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

3 x^{2}

‖

\frac{d}{d x} (- 12 x)

-12

⇥

3 x^{2} - 12

⇥The first derivative is

3 x^{2} - 12

⇥1.2) Let's solve in

ℝ

3 x^{2} - 12 = 0

⇥You can write the expression in factored form:

3 \cdot (x^{2} - 4) = 0

⇥Let's solve:

x^{2} - 4 = 0

⇥Solutions obtained:

x = {-2, 2}

⇥1.3) Let's differentiate:

\frac{d}{d x} (3 x^{2} - 12)

⇥

\frac{d}{d x} (3 x^{2}) + 0

⇥ℹ

\frac{d}{d x} (3 x^{2})

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

‖

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

2 x

⇥

3 \cdot 2 x

⇥

6 x

⇥The second derivative is

6 x

2) At

x = -2

, calculate

f'' (-2) = 6 \cdot (-2) = -12

x = -2

, the second derivative is negative: this is a strict local maximum

⇥

{(-2)}^{3} - 12 \cdot (-2) + 2

⇥ℹ

{(-2)}^{3}

-8

‖

-12 \cdot (-2)

24

⇥

- 8 + 24 + 2

⇥

16 + 2

⇥

18

The y-coordinate is f(

-2

) =

18

3) At

x = 2

, calculate

f'' (2) = 6 \cdot 2 = 12

x = 2

, the second derivative is positive: this is a strict local minimum

⇥

2^{3} - 12 \cdot 2 + 2

⇥ℹ

2^{3}

8

‖

-12 \cdot 2

-24

⇥

8 - 24 + 2

⇥

- 16 + 2

⇥

-14

The y-coordinate is f(

2

) =

-14

This odd-degree polynomial is unbounded above and below on ℝ: it has no global extremum

▶Strict local extrema:

{(-2, 18), (2, -14)}