critical 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: ddx(x3 -6x2 + 9x + 1)
⇥ⓘ The derivative of a sum is the sum of the derivatives: ddx(x3 -6x2 + 9x + 1) = ddx(x3) + ddx(-6x2) + ddx(9x) + 0
⇥ ddx(x3) + ddx(-6x2) + ddx(9x) + 0
⇥ⓘ The derivative of x^n is n*x^(n-1): ddx(x3) = 3x2, ddx(x2) = 2x■A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(-6x2) = -6·ddx(x2), ddx(9x) = 9■Adding 0 changes nothing: ddx(9x) + 0 = ddx(9x)
⇥ 3x2 -6·2x + 9
⇥ⓘ Multiplications are done before additions and subtractions: 6·2 = 12
⇥ 3x2 -12x + 9
⇥1.2) Let's solve in ℝ: 3x2 -12x + 9 = 0
⇥Since 3 ≠ 0, divide both sides by 3: x2 -4x + 3 = 0
⇥Given the quadratic form ax2 + bx + c, let's compute the discriminant: Δ = b2 -4ac. Here a = 1, b = -4 and c = 3
⇥ Δ = (-4)2 -4·1·3
⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: (-4)2 = (-4)·(-4) = 16
⇥ Δ = 16 -4·3
⇥ⓘ Multiplications are done before additions and subtractions: 4·3 = 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 x1 = -b -Δ2a = -(-4) -42·1
⇥ x1 = 4 -42
⇥ⓘ The square root of a number is the positive number whose square is that number: 4 = 2; 22 = 4
⇥ x1 = 4 -22
⇥ⓘ The numerator and the denominator are computed before the division: 4 -2 = 2
⇥ x1 = 22
⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by 2, their greatest common divisor: 22 = 1
⇥ x1 = 1
⇥The quadratic formula gives x2 = -b + Δ2a = -(-4) + 42·1
⇥ x2 = 4 + 42
⇥ⓘ The square root of a number is the positive number whose square is that number: 4 = 2; 22 = 4
⇥ x2 = 4 + 22
⇥ⓘ The numerator and the denominator are computed before the division: 4 + 2 = 6
⇥ x2 = 62
⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by 2, their greatest common divisor: 62 = 3
⇥ x2 = 3
▶Solutions obtained: x = { 1, 3 }