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]
Use the product rule to get
f ' (t) = -(t^2 - 32)/√(64-t^2)
= 0 for a local max/min
t^2 = 32
t = ±4√2
f(±4√2)
= 0
f(-1) = -1√63
f(8) = 8(√0) = 0
there you go, you got your f(t)'s
what do you think?
Here is your graph, notice it only defined between -8 and +8
http://www.wolframalpha.com/input/?i=y+%3D+++t+√%2864+−+t%5E2%29
To find the absolute maximum and absolute minimum values of the function f(t) = t√(64 - t^2) on the interval [-1, 8], we need to follow these steps:
Step 1: Find the critical points of the function within the interval [-1, 8]. These are the points where the derivative of the function is either zero or undefined.
Step 2: Evaluate the function at the critical points and the endpoints of the interval.
Step 3: Compare the values obtained in step 2 to determine the absolute maximum and absolute minimum values.
Let's proceed with each step:
Step 1: Find the critical points
To find the critical points, we need to find where the first derivative of the function is zero or undefined.
First, let's find the derivative of f(t):
f'(t) = d/dt [t√(64 - t^2)]
To simplify this derivative, we can use the product rule:
f'(t) = (t) * (d/dt √(64 - t^2)) + (√(64 - t^2)) * (d/dt t)
Taking the derivatives, we get:
f'(t) = √(64 - t^2) + (t * (-1/2) * (2t) / √(64 - t^2))
f'(t) = √(64 - t^2) + (-t^2 / √(64 - t^2))
To find the critical points, we need to solve f'(t) = 0:
√(64 - t^2) + (-t^2 / √(64 - t^2)) = 0
Now, let's simplify this equation further:
√(64 - t^2) - (t^2 / √(64 - t^2)) = 0
√(64 - t^2) = (t^2 / √(64 - t^2))
Squaring both sides of the equation:
64 - t^2 = t^4 / (64 - t^2)
Cross-multiplying:
(64 - t^2)^2 = t^4
Expanding and rearranging the equation:
4096 - 128t^2 + t^4 - t^4 = 0
128t^2 = 4096
t^2 = 4096 / 128
t^2 = 32
Taking the square root of both sides:
t = ±√32
t = ±4√2
Therefore, the critical points of the function f(t) within the interval [-1, 8] are t = -4√2 and t = 4√2.
Step 2: Evaluate the function at the critical points and endpoints
Now, we need to evaluate the function f(t) = t√(64 - t^2) at the critical points and endpoints of the interval [-1, 8].
f(-1) = (-1)√(64 - (-1)^2) = -√63
f(8) = (8)√(64 - 8^2) = 8√0 = 0
f(-4√2) = (-4√2)√(64 - (-4√2)^2) = -4√8
f(4√2) = (4√2)√(64 - (4√2)^2) = 4√8
Step 3: Compare the values
Now, let's compare the values obtained in step 2 to determine the absolute maximum and absolute minimum.
The function values are as follows:
f(-1) = -√63
f(8) = 0
f(-4√2) = -4√8
f(4√2) = 4√8
From these values, we can see that:
The absolute maximum value of f(t) on the interval [-1, 8] is 4√8, which occurs at t = 4√2.
The absolute minimum value of f(t) on the interval [-1, 8] is -√63, which occurs at t = -1.
Therefore, the absolute maximum and absolute minimum values of the function f(t) on the interval [-1, 8] are 4√8 and -√63, respectively.