Let f(t)=t\sqrt{4-t} on the interval [-1,3]. Find the absolute maximum and absolute minimum of f(t) on this interval.
The absolute max occurs at t=.
The absolute min occurs at t=
To find the absolute maximum and absolute minimum of a function on a closed interval, we can use the Extreme Value Theorem and calculate the critical points and endpoints of the interval.
1. Calculate the derivative of f(t):
f'(t) = (d/dt) (t√(4-t))
Using the product rule, we get:
f'(t) = √(4-t) - (t/(2√(4-t)))
2. Find the critical points:
To find the critical points, we need to find where the derivative is equal to zero or undefined.
Setting f'(t) = 0, we have:
√(4-t) - (t/(2√(4-t))) = 0
Multiply through by 2√(4-t) to eliminate the fractions:
2(4-t) - t = 0
8 - 2t - t = 0
8 - 3t = 0
3t = 8
t = 8/3
3. Calculate the endpoints of the interval:
The endpoints of the interval [-1,3] are -1 and 3.
4. Evaluate the function f(t) at the critical points and endpoints:
f(-1) = (-1)√(4-(-1)) = (-1)√(5) = -√5
f(3) = (3)√(4-3) = (3)√(1) = 3
f(8/3) = (8/3)√(4-(8/3))
To simplify further, we can rationalize the denominator:
f(8/3) = (8/3)√(12/3 - 8/3) = (8/3)√(4/3) = (8/3)(2/√3) = 16/√3
5. Compare the values of f(t) at the critical points and endpoints:
The absolute maximum occurs at the endpoint t = 3 with a value of f(3) = 3.
The absolute minimum occurs either at the endpoint t = -1 with a value of f(-1) = -√5 or at the critical point t = 8/3 with a value of f(8/3) = 16/√3.
Therefore, the absolute maximum of f(t) on the interval [-1,3] occurs at t = 3, and the absolute minimum occurs at t = -1 or t = 8/3.