Final answer:
The absolute minimum value of the function f(x) = 12x - 203 on the closed interval [0,3] is found by evaluating the function at the endpoints of the interval, as the function's derivative reveals no critical points. The function takes on its minimum value at x = 0, which is -203.
Step-by-step explanation:
The question asks to find the absolute minimum value of the function f(x) = 12x - 203 on the closed interval [0,3]. To find the minimum value of a function over a closed interval, we need to evaluate the function at the critical points and the endpoints of the interval.
First, let's find the derivative of the function to determine if there are any critical points inside the interval. The derivative of f(x) is f'(x) = 12, which is a constant. This means there are no critical points because the derivative never equals zero.
Now we need to evaluate the function at the endpoints of the interval, which are x = 0 and x = 3.
- For x = 0, we have f(0) = 12(0) - 203 = -203.
- For x = 3, we have f(3) = 12(3) - 203 = 36 - 203 = -167.
Comparing these two values, the absolute minimum value of f(x) on the interval [0,3] is -203 since -203 < -167.