critical f(x)=x^3-12x+6 — step-by-step solution

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

⇥Let's perform the differentiation:

\frac{d}{d x} (x^{3} - 12 x + 6)

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

\frac{d}{d x} (x^{3} - 12 x + 6)

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

⇥

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

⇥ⓘ The derivative of x^n is n*x^(n-1):

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

3 x^{2}

■Adding 0 changes nothing:

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

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

■A constant factor stays in front of the derivative, (k*u)' = k*u':

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

-12

⇥

3 x^{2} - 12

⇥Let's solve in

ℝ

3 x^{2} - 12 = 0

⇥ⓘ Both sides are divided by

3

⇥

x^{2} = 4

⇥The two numbers whose square is

4

are

-2

and

2

▶Solutions obtained:

x = {-2, 2}