Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = (36 − t^2)^ 1/t, [−1, 6]
nasty equation, none of the standard methods of derivatives fit.
I am going to take logs of both sides
ln y = ln (36- t^2)^(1/t)
= (1/t)(ln (36-t^2)
now product rule on right side
(dy/dt) / y = (1/t)(-2t)/(36-t^2) + (-1/t^2)(ln(36-t^2) )
dy/dt = y[(1/t)(-2t)/(36-t^2) + (-1/t^2)(ln(36-t^2) )]
so y = 0
(36-t^2)^(1/t) = 0
there is an intuitive solution of t = 6
or
(1/t)(-2t)/(36-t^2) + (-1/t^2)(ln(36-t^2) ) = 0
-2/(36-t^2) = ln(36-t^2) /t^2
ln(36-t^2) = -2t^2/(36-t^2)
what a messy equation, ran it through Wolfram
and there are no real solutions, (4 complex)
http://www.wolframalpha.com/input/?i=solve++log%2836-t%5E2%29+%3D+-2t%5E2%2F%2836-t%5E2%29
unless I made an algebraic error or typo
so evaluate f(6) which happens to be at the end of your interval.
f(6) = (36-36)^(1/6 = 0
Also evalute f(-1) to see which is the max or min.
f(-1) = (36 - 1)^-1
=1/35 = appr .02857
To find the absolute maximum and absolute minimum values of the function f(t) = (36 - t^2)^(1/t) on the interval [-1, 6], we can follow these steps:
Step 1: Find the critical points of f(t) within the interval [-1, 6]. Critical points are the values of t where the derivative of f(t) is either zero or undefined.
Step 2: Evaluate f(t) at the critical points and the endpoints of the interval [-1, 6].
Step 3: Compare the values obtained in Step 2 to determine the absolute maximum and minimum values.
Let's begin with Step 1 by finding the critical points:
1. Find the derivative of f(t):
To find the derivative of f(t), use the chain rule, which states that d(u^v)/dt = v(u^(v-1))(du/dt) + (ln(u))(u^v)(dv/dt).
Applying the chain rule to f(t) = (36 - t^2)^(1/t), we have:
f'(t) = (1/t)((36 - t^2)^(1/t-1))(d(36 - t^2)/dt)
Now, we need to find d(36 - t^2)/dt:
d(36 - t^2)/dt = -2t
Plugging it back into f'(t), we get:
f'(t) = (1/t)((36 - t^2)^(1/t-1))(-2t)
2. Set f'(t) = 0 and solve for t:
(1/t)((36 - t^2)^(1/t-1))(-2t) = 0
Since division by zero is not allowed, we can set the numerator equal to zero:
(36 - t^2)^(1/t-1) = 0
Since the expression (36 - t^2)^(1/t-1) is always positive (as it involves an exponentiation), it can never be equal to zero. Hence, there are no critical points when f'(t) = 0 within the given interval.
Now, let's proceed to Step 2 and evaluate f(t) at the critical points and the endpoints:
Evaluate f(t) at the endpoints:
f(-1) = (36 - (-1)^2)^(1/(-1)) = (36 - 1)^(1/(-1)) = 35^(-1) = 1/35
f(6) = (36 - 6^2)^(1/6) = (36 - 36)^(1/6) = 0^(1/6) = 0
Since there are no critical points, we only need to check the endpoints to find the absolute maximum and minimum values.
Step 3: Comparing the values obtained in Step 2:
The absolute maximum value of f(t) on the interval [-1, 6] is 1/35, which occurs at t = -1.
The absolute minimum value of f(t) on the interval [-1, 6] is 0, which occurs at t = 6.
Therefore, the absolute maximum value is 1/35 and the absolute minimum value is 0.