Answer:
Below in bold.
Explanation:
f(x)=2x^3-3x^2
Finding the derivative and equating to zero:
f'(x) = 6x^2 - 6x = 0
6x(x - 1) = 0
So there are turning point at x = 0 and 2 = 1.
Finding the relative maximum:
Second derivative is negative when we have a maximum value:
f"(x) = 12x - 5
- which is negative when x = 0.
So the relative maximum is at x = 0 where f(x) = 2(0)^3 - 3(0)^2 = 0.
At x = 1, second derivative = 12(1) - 5 = +7 so this is a relative minimum.
The function is increasing in interval -∝ < x < 0
decreasing in interval 0 < x < 1
then increasing in interval 1 < x < ∝