Investigating Polynomial Functions Worksheet

What is the derivative of f(x) = x³ - 12x + 5?

f'(x) = 3x² - 12

Find the critical x-values for f(x) = x³ - 12x + 5 by setting the derivative to zero.

3x² - 12 = 0 leads to x² = 4, so x = 2 and x = -2.

Find the full (x, y) coordinates of the stationary points for f(x) = x³ - 12x + 5.

f(2) = (2)³ - 12(2) + 5 = -11, giving (2, -11). f(-2) = (-2)³ - 12(-2) + 5 = 21, giving (-2, 21).

Using a sign table, classify each stationary point of f(x) = x³ - 12x + 5.

f'(-3) = 15 (>0), f'(0) = -12 (<0), f'(3) = 15 (>0). At x = -2, f' changes from + to -, so (-2, 21) is a local maximum. At x = 2, f' changes from - to +, so (2, -11) is a local minimum.

State the intervals on which f(x) = x³ - 12x + 5 is increasing and decreasing.

Increasing on x < -2 and x > 2; decreasing on -2 < x < 2.

For g(x) = x⁴ - 4x³, find the derivative and all stationary points.

g'(x) = 4x³ - 12x² = 4x²(x - 3). Setting g'(x) = 0 yields x = 0 and x = 3. Coordinates are (0, 0) and (3, -27).

Classify the stationary point at (0, 0) for g(x) = x⁴ - 4x³ using test points in g'(x).

For x < 0, say x = -1: g'(-1) = 4(1)(-4) = -16 (<0). For 0 < x < 3, say x = 1: g'(1) = 4(1)(-2) = -8 (<0). Since the sign does not change, (0, 0) is a horizontal inflection point, not an extremum.