critical f(x)=3x^2-130x+1000 — 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} (3 x^{2} - 130 x + 1000)

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

\frac{d}{d x} (3 x^{2} - 130 x + 1000)

\frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 130 x) + 0

⇥

\frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 130 x) + 0

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

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

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

\frac{d}{d x} (- 130 x)

-130

■The derivative of x^n is n*x^(n-1):

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

2 x

■Adding 0 changes nothing:

\frac{d}{d x} (- 130 x) + 0

\frac{d}{d x} (- 130 x)

⇥

3 \cdot 2 x - 130

⇥ⓘ Multiplications are done before additions and subtractions:

3 \cdot 2

6

⇥

6 x - 130

⇥Let's solve in

ℝ

6 x - 130 = 0

⇥ⓘ We add

130

to both sides

⇥

6 x = 130

⇥ⓘ Both sides are divided by

6

⇥

x = \frac{130}{6}

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

2

, their greatest common divisor:

\frac{130}{6}

\frac{65 \cdot 2}{3 \cdot 2}

\frac{65}{3}

▶Solution found:

x = \frac{65}{3}