Find the intervals on which f is increasing and the intervals on which it is decreasing.

Question

f(x)= x^4/4 - 3x^3 + 9x^2

Answer 1

I will first solve this without using Calculus.

x^4/4 - 3x^3 +9x^2 = x^2/4 (x^2 - 12x + 36) = x^2/4 (x - 6)^2

Thus, we see that we have two global minima at x = 0 and x = 6, and the function will be increasing for x > 6 and decreasing for x < 0 (going to infinity as x goes to - infinity).

We know there must be a relative maximum between 0 and 6, and our "guess" is that it is at 3. We will show this.

We can rewrite x^2/4 (x - 6)^2 = (x (x-6))^2 = (x^2 - 6x)^2 = ((x-3)^2 - 9)^2 = (9 - (x-3)^2)^2

Clearly, the term in parentheses is a parabola with a positive max 9 at x = 3 and is positive with no other critical points on (0, 6) and reaches 0 at [0,6]. Thus, the function itself has a maximum at 3 and no other critical points on (0, 6)

Thus, we have a relative maximum at 3 and 2 relative minima at 0 and 6.

The function increases on (0, 3) and (6, ∞) and decreases on (-∞, 0) and (3, 6)

Now, to use Calculus.

f'(x) = x^3 - 9x^2 + 18x = x(x - 3)(x - 6)

f''(x) = 3x^2 - 18 x + 18

f''(0) = 18, so 0 is a min.

f''(3) = 3 * 3^2 - 18*3 + 18 = -9, so 3 is a maximum.

f''(6) = 3*6^2 - 18 * 6 + 18 = 18, so 6 is a minimum. (As expected, f''(6) = f''(0))

Then, the function increases on (0, 3) and (6, ∞) and decreases on (-∞, 0) and (3, 6)