Question

Find the absolute extrema if they exist, as well as all values of x where they occur, for the function f(x)=2x-3x²/3. (a) on the interval [-1,0.5] and (b) on the interval [0.5,2].

Answer 1

To find the absolute extrema of the function f(x) = 2x - (3x^2)/3 on the given intervals, we need to find the critical points and evaluate the function at the endpoints.

Step-by-step explanation:

To find the absolute extrema of the function f(x) = 2x - (3x^2)/3 on the interval [-1, 0.5], we need to find the critical points and endpoints. First, let's find the critical points by taking the derivative of f(x) and setting it equal to zero:

f'(x) = 2 - 2x = 0

Solving for x, we get x = 1. Next, we evaluate f(x) at the critical point and the endpoints:

f(-1) = 2(-1) - (3(-1)^2)/3 = -2 + 1 = -1

f(0.5) = 2(0.5) - (3(0.5)^2)/3 = 1 - 0.25 = 0.75

Since f(-1) = -1, f(0.5) = 0.75, and f(1) = 1, the maximum value is 1 and it occurs at x = 1. The minimum value is -1 and it occurs at x = -1.

To find the absolute extrema on the interval [0.5, 2], we repeat the same steps. The critical point is still x = 1, but now we need to evaluate f(x) at the endpoints:

f(0.5) = 0.75

f(2) = 2(2) - (3(2)^2)/3 = 4 - 4 = 0

Since f(0.5) = 0.75, f(1) = 1, and f(2) = 0, the maximum value is 1 and it occurs at x = 1. The minimum value is 0 and it occurs at x = 2.