Absolute Maximum & Minimum Values?

Question

Find the absolute maximum & minimum values of f(t)=2t(16-t^2)^(1/2) in the interval [-2, 4]. Give the answer in the form of f(a1)=b1, f(a2)=b2. Do not round the answers.

Answer 1

Well first I tried to find the derivative of your function, and I think it is:

f '(t) = - 2 t^2 (16 - t^2)^(-1/2) + 2(16 - t^2)^(1/2)

From there, I tried to find the value where the derivative is equal to zero.

Using my handheld calculator I found that:

f '(t) = 0 when t= 2.82842712475.

So you have a maximum value : f(2.82842712475)=16

And also, I would GUESS that the derivative will not reveal a minima until t=-2.82842712475. Because of the square root, we must follow our instincts: the calculator is not able to go there.

So the minima will be found at the lower limit of your interval which is -2. Therefore,

f(-2) = -13.8564064606

This solution may still contain mistakes or errors, if so they were not intended. Good Luck.

Answer 2

F(-2)=-4*sqrt(12) is the minimum in the interval specified.
finding the other answer is a problem in differential calculus
Take the first derivative of the function and find a zero.
The answer will be something between t=2.5 and t=3.