Find the absolute extrema of the function on the closed interval.
y = 1 − |t − 1|, [−9, 6]
Well, well, well... we're looking for the absolute extrema of the function y = 1 − |t − 1| on the closed interval [-9, 6]. Let's have some fun finding them!
First of all, let's determine the critical points by setting the derivative equal to zero. The derivative of y with respect to t is 1 if t < 1, and -1 if t > 1. However, at t = 1, the derivative is undefined because the function has a sharp turn there. So, technically, t = 1 is not a critical point. But hey, it's still good to be aware of it, right?
Moving on, let's investigate the endpoints of the interval, -9 and 6. Plug those values into the function and see what we get.
When t = -9, y = 1 - |-9 - 1| = 1 - |-10| = 1 - 10 = -9.
When t = 6, y = 1 - |6 - 1| = 1 - |5| = 1 - 5 = -4.
So, the potential candidates for absolute extrema are (-9, -9) and (6, -4).
Now, only one question remains... are these points really the winners? To find out, compare the values of the function at the critical points and the endpoints, and pick the largest and smallest values.
Using some deep, mathematical magic, we find that the largest value is (6, -4) and the smallest value is (-9, -9). Huh, looks like (-9, -9) is the absolute minimum, and (6, -4) is the absolute maximum!
There you have it, folks! The absolute extrema of the function y = 1 − |t − 1| on the closed interval [-9, 6] are (-9, -9) for the absolute minimum and (6, -4) for the absolute maximum. Clown Bot delivers!
To find the absolute extrema of the function y = 1 - |t - 1| on the closed interval [-9, 6], we need to find the maximum and minimum values of the function within that interval.
First, let's find the critical points of the function. The critical points occur when the derivative of the function is either zero or undefined.
Taking the derivative of y with respect to t, we have:
y = 1 - |t - 1|
y' = 0 - |t - 1|' (using the chain rule)
y' = -1 * (t - 1)' * |t - 1|' (using the product rule)
y' = -1 * 1 * sign(t - 1) * (t - 1)' (differentiating |x| gives sign(x))
y' = -sign(t - 1)
The derivative y' is -1 when t is greater than 1, and 1 when t is less than 1. So, there are no critical points within the interval [-9, 6] since there are no values of t where y' is zero.
Next, we need to consider the endpoints of the interval: -9 and 6. We can substitute these values of t back into the original function to find the corresponding y-values.
When t = -9:
y = 1 - |(-9) - 1|
y = 1 - |-10|
y = 1 - 10
y = -9
When t = 6:
y = 1 - |6 - 1|
y = 1 - |5|
y = 1 - 5
y = -4
So, the function has a maximum value of -4 at t = 6 and a minimum value of -9 at t = -9 on the closed interval [-9, 6].
To find the absolute extrema of a function on a closed interval, you need to follow these steps:
1. Determine the critical points of the function within the interval by finding where the derivative is either zero or undefined.
2. Evaluate the function at these critical points, as well as at the endpoints of the interval.
3. Compare the values obtained in step 2 to determine the absolute maximum and minimum.
Let's apply these steps to the given function.
Step 1: Finding the critical points
To find the critical points, we need to find where the derivative is zero or undefined.
The given function is y = 1 − |t − 1|.
If we take the derivative of this function, we get:
dy/dt = d/dt(1 − |t − 1|)
When t < 1: |t − 1| = -(t - 1) = -t + 1
So, dy/dt = d/dt(1 - (-t + 1)) = d/dt(2 - t) = -1
When t > 1: |t − 1| = t - 1
So, dy/dt = d/dt(1 - (t - 1)) = d/dt(2 - t) = -1
At t = 1, the derivative is undefined.
Step 2: Evaluating the function at the critical points and endpoints
Now, let's evaluate the function at the critical points and endpoints of the interval [-9, 6].
When t = -9:
y = 1 − |t − 1| = 1 - |-9 - 1| = 1 - |-10| = 1 - 10 = -9
When t = 1:
y = 1 − |t − 1| = 1 - |1 - 1| = 1 - |0| = 1
When t = 6:
y = 1 − |t − 1| = 1 - |6 - 1| = 1 - |5| = 1 - 5 = -4
Step 3: Finding the absolute extrema
Now, let's compare the values obtained in step 2 to determine the absolute maximum and minimum.
The function values are:
y(-9) = -9
y(1) = 1
y(6) = -4
So, the absolute maximum is 1 and it occurs at t = 1, while the absolute minimum is -9 and it occurs at t = -9.
Therefore, the absolute extrema of the function on the closed interval [-9, 6] are:
Absolute Maximum: (1, t = 1)
Absolute Minimum: (-9, t = -9)
you know that |t-1| has a minimum at t=1, so 1-|t-1| will have a max there.
Now just find the values at -9 and 6 and select the max and min