Derivative of a function

Question

Suppose the derivative of a function f is f '(x) = (x + 1)⁴(x − 3)⁵(x − 6)⁶.

 On what interval is f increasing? (Enter your answer in interval notation.)

 

Answer 1

f(x) increases where f'(x)>0:

f'(x) = (x+1)⁴(x-3)⁵(x-6)⁶

0 = (x+1)⁴(x-3)⁵(x-6)⁶

x = {-1,3,6}

Use test points between each critical point:

f'(-2) = (-2+1)⁴(-2-3)⁵(-2-6)⁶ = (-1)⁴(-5)⁵(-8)⁶ < 0 (negative)

f'(0) = (0+1)⁴(0-3)⁵(0-6)⁶ = (1)⁴(3)⁵(-6)⁶ > 0 (positive)

f'(4) = (4+1)⁴(4-3)⁵(4-6)⁶ = (5)⁴(1)⁵(-2)⁶ > 0 (positive)

f'(7) = (7+1)⁴(7-3)⁵(7-6)⁶ = (8)⁴(4)⁵(1)⁶ > 0 (positive)

As we can see, (-∞,-1] is where the function f(x) decreases since f'(x)<0 there, and (-1,3) ∪ (3,6) ∪ (6,∞) is where f(x) increases since f'(x)>0 over those intervals.

Hope this helped!