Question

Find the absolute​ minimum, if it exists. f(x)= (x-2)^2 if 1≤x<3
and f(x)= -3x+16 if 3≤x<7

Answer 1

To find the absolute minimum of the function f(x), we need to check the critical points and endpoints of the intervals.

First, let's find the critical point within the interval 1 ≤ x < 3. We take the derivative of the function f(x) within this interval:
f'(x) = 2(x - 2)

Setting f'(x) equal to 0:
2(x - 2) = 0
x - 2 = 0
x = 2

Since 2 is within the interval 1 ≤ x < 3, we need to check the function value at x = 2:
f(2) = (2 - 2)^2
f(2) = 0

Next, within the interval 3 ≤ x < 7, we need to find the critical point:
For this interval, f(x) = -3x + 16, taking the derivative:
f'(x) = -3

Setting f'(x) equal to 0:
-3 = 0

There is no critical point within this interval.

Now, we need to check the endpoints of the intervals:

1. When x = 1:
f(1) = (1 - 2)^2
f(1) = 1

2. When x = 3:
f(3) = -3(3) + 16
f(3) = 7

To find the absolute minimum value, we compare the values at the critical point, endpoints, and wherever the function changes between the intervals.
The minimum value is at x = 2, where f(2) = 0.

Therefore, the absolute minimum value of the function f(x) is 0.