Find the absolute min and max values for f(t) = t*sqrt(36-t^2) on the interval [-1,6]

Question

anonymous · Answer

you need to find the derivative of f(t) first.  and find the values that make the derivative function 0. Then also plug in 6 and -1 into the equation to see if the endpoints are max or min values.

anonymous · Answer

I've attempted just that. i got a min value of t = -1 and max val of t = sqrt(18), but those don't appear to be the answers

anonymous · Answer

what did you get for the zeros of the derivative?

anonymous · Answer

sqrt(18) and + or - 6

anonymous · Answer

f(3)=3*sqrt(1)=3*1=3   ab max: (8/3, 3.08)     ab min: (-1, -2.24)

anonymous · Answer

where is the 3 coming from?