Answer:
the function f(x) = 2x^2 - 3x is strictly decreasing on the interval (-∞, 3/4).
Explanation:
To find the intervals in which the function f(x) = 2x^2 - 3x is strictly increasing or strictly decreasing, we need to find the first derivative of the function and then determine the sign of the derivative over different intervals.
(a) To find the intervals in which the function f(x) = 2x^2 - 3x is strictly increasing, we need to find where the first derivative is positive. The first derivative of f(x) is:
f'(x) = 4x - 3
To determine the sign of f'(x), we set it equal to zero and solve for x:
4x - 3 = 0
4x = 3
x = 3/4
This critical point divides the real number line into two intervals: (-∞, 3/4) and (3/4, ∞).
To determine the sign of f'(x) over each interval, we can pick a test point in each interval and plug it into the derivative. For example, if we choose x = 0, we have:
f'(0) = 4(0) - 3 = -3
Since f'(0) is negative, we know that f(x) is decreasing on the interval (-∞, 3/4).
If we choose x = 1, we have:
f'(1) = 4(1) - 3 = 1
Since f'(1) is positive, we know that f(x) is increasing on the interval (3/4, ∞).
Therefore, the function f(x) = 2x^2 - 3x is strictly increasing on the interval (3/4, ∞).
(b) To find the intervals in which the function f(x) = 2x^2 - 3x is strictly decreasing, we need to find where the first derivative is negative. Using the same process as above, we find that f'(x) = 4x - 3 and the critical point is x = 3/4.
Picking test points in the intervals (-∞, 3/4) and (3/4, ∞), we find that f(x) is strictly decreasing on the interval (-∞, 3/4).
Therefore, the function f(x) = 2x^2 - 3x is strictly decreasing on the interval (-∞, 3/4).