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MatroZZZ [7]
3 years ago
14

Let f(x)=x^3-3x^2+3.

Mathematics
1 answer:
rusak2 [61]3 years ago
5 0

Answer:

a) Domain: all real values of x

Range: all real values of y

b) Relative maxima: (0,3)

Relative minima: (2,-1)

c) yes, (1,1)

d) x < 0, x > 2

Step-by-step explanation:

a) Domain: all real values of x

Range: all real values of y

b) dy/dx = 3x² - 6x = 0

3x(x - 2) = 0

x = 0, y = 0³ - 3(0)² + 3 = 3

x = 2, y = 2³ - 3(2)² + 3 = -1

d²y/dx² = 6x - 6

x = 0, d²y/dx² = -6 < 0 (max)

x = 2, d²y/dx² = 6 > 0 (min)

c) d²y/dx² = 0 at points of infection

6x - 6 = 0

x = 1

y = 1³ - 3(1)² + 3 = 1

dy/dx at x = 0.9: 3(0.9)² - 6-(0.9)

= -0.298

dy/dx at x = 1.1: 3(1.1²) - 6(1.1)

= -0.297

Decreasing on both sides, therefore (1,1) is a point of inflection

d) increasing when dy/dx > 0

3x² - 6x > 0

Roots: x = 0, 2

x < 0, x > 2

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