critical x^3-48x — step-by-step solution

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

⇥Let's differentiate:

\frac{d}{d x} (x^{3} - 48 x)

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

\frac{d}{d x} (x^{3} - 48 x)

\frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 48 x)

⇥

\frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 48 x)

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

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

3 x^{2}

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

\frac{d}{d x} (- 48 x)

-48

⇥

3 x^{2} - 48

⇥Let's solve in

ℝ

3 x^{2} - 48 = 0

⇥ⓘ Both sides are divided by

3

⇥

x^{2} = 16

⇥The two numbers whose square is

16

are

-4

and

4

▶Solutions found:

x = {-4, 4}