Answer:
On (-2/3, 1) the given function is decreasing
Step-by-step explanation:
Do you know calculus? If so:
Differentiate y = 2x^3 – x^2 – 4x + 5 and set the derivative = to 0:
dy/dx = 6x^2 - 2x - 4 = 0, or 3x^2 - x - 2 = 0.
This factors as follows: (x - 1)(3x + 2) = 0.
The roots of this equation are {1, -2/3}.
Plot these two roots on a number line and then set up intervals as follows:
(-infinity, -2/3), (-2/3, 1), (1, infinity)
Choose a test number from each interval: { -1, 0, 2 }
By evaluating the derivative 6x^2 - 2x - 4 at each of these three test numbers, we get:
dy/dx = 6x^2 - 2x - 4 is positive on (-infinity, -2/3), and so we conclude that the given function is increasing on that interval.
dy/dx = 6x^2 - 2x - 4 is negative on (-2/3, 1), and so we conclude that the given function is decreasing on that interval. To the nearest tenth:
(-0.7, 1)
