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
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]
1 answer
1 answer