Locate the absolute exterma of the function on the closed interval.
y=3-|t-3| , [-1,5]
Well, |t-3| >= 0, becoming its minimum value (0) at t=3.
So, y has a maximum value of 3 at t=3, since it is subtracting 0 from 3. For anything other than 3, y is less than 3.
The other extrema would be minima, occurring at the ends of the interval.
To locate the absolute extrema of the function y = 3 - |t - 3| on the closed interval [-1, 5], we need to evaluate the function at the critical points and endpoints within the interval.
1. Find the critical points:
To find the critical points, we need to locate the points where the derivative of the function is either zero or undefined.
Taking the derivative of y with respect to t:
y = 3 - |t - 3|
y' = -1 * sign(t - 3) * 1
The derivative is undefined at t = 3.
2. Evaluate the function at the endpoint t = -1:
y(-1) = 3 - |-1 - 3|
y(-1) = 3 - |-4|
y(-1) = 3 - 4
y(-1) = -1
3. Evaluate the function at the endpoint t = 5:
y(5) = 3 - |5 - 3|
y(5) = 3 - |2|
y(5) = 3 - 2
y(5) = 1
4. Evaluate the function at the critical point t = 3:
At t = 3, the function y is defined as y(3) = 3 - |3 - 3|
y(3) = 3 - |0|
y(3) = 3 - 0
y(3) = 3
Now, we compare these values to determine the absolute extrema:
The function has a local minimum at t = -1 with a value of y = -1.
The function has a local maximum at t = 3 with a value of y = 3.
The function has a local minimum at t = 5 with a value of y = 1.
Therefore, the absolute minimum value of the function y = 3 - |t - 3| on the closed interval [-1, 5] is -1, and the absolute maximum value is 3.
To locate the absolute extrema of the function y = 3 - |t - 3| on the closed interval [-1, 5], we need to find the highest and lowest points on that interval.
To do this, we can follow these steps:
1. Calculate the function values at the endpoints of the interval: Plug in t = -1 and t = 5 into the function y = 3 - |t - 3| to find the y-values at those points.
- When t = -1: y = 3 - |-1 - 3| = 3 - |-4| = 3 - 4 = -1
- When t = 5: y = 3 - |5 - 3| = 3 - |2| = 3 - 2 = 1
2. Determine the critical points of the function: Critical points occur where the derivative of the function is either zero or does not exist. In this case, the function y = 3 - |t - 3| has a sharp point at t = 3, where the absolute value changes direction.
- Take the derivative of y = 3 - |t - 3| with respect to t:
- When t < 3: y' = 0
- When t > 3: y' = 0
3. Calculate the function value at the critical point: Plug in t = 3 into the function y = 3 - |t - 3| to find the y-value at that point.
- When t = 3: y = 3 - |3 - 3| = 3 - |0| = 3 - 0 = 3
4. Compare the y-values obtained in steps 1, 2, and 3 to determine the absolute extrema:
- The lowest y-value is -1, which occurs at t = -1.
- The highest y-value is 3, which occurs at t = 3.
Therefore, the absolute minimum is -1 and occurs at t = -1, while the absolute maximum is 3 and occurs at t = 3.