Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = (t square root of (64 − t^2)), [−1, 8]
so it the same way you were just given in your previous post.
Let us know what you got
To find the absolute maximum and minimum values of a function on a given interval, we need to follow these steps:
1. Find the critical points of the function.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. Identify the largest and smallest values to determine the absolute maximum and minimum.
Let's apply these steps to the given function f(t) = t√(64 − t^2), on the interval [−1, 8].
Step 1: Find the critical points.
To find the critical points, we need to find the values of t where the derivative of the function is equal to zero or does not exist.
Firstly, let's find the derivative of f(t):
f'(t) = √(64 - t^2) + t(1/2)(64 - t^2)^(-1/2)(-2t)
= √(64 - t^2) - (t^2)/(√(64 - t^2))
= (64 - 2t^2)/(√(64 - t^2))
The derivative f'(t) will be zero when the numerator is zero, i.e.,
64 - 2t^2 = 0
Solving this equation, we get:
2t^2 = 64
t^2 = 32
t = ±√32
So, the critical points are t = √32 and t = −√32.
Step 2: Evaluate the function at the critical points and endpoints.
Now we need to evaluate the function at t = -1, t = 8, t = √32, and t = -√32.
For t = -1:
f(-1) = (-1)√(64 - (-1)^2) = -√63
For t = 8:
f(8) = (8)√(64 - 8^2) = 0
For t = √32:
f(√32) = (√32)√(64 - (√32)^2) = √(32 * 32) = 32
For t = -√32:
f(-√32) = (-√32)√(64 - (-√32)^2) = -√(32 * 32) = -32
Step 3: Determine the absolute maximum and minimum.
Now we compare the function values to find the largest and smallest values.
The maximum value is 32 at t = √32.
The minimum value is -√63 at t = -1.
So, the absolute maximum value is 32, and the absolute minimum value is -√63.