SOLUTION: Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x)=12x^(2/3) -x on [0,1728]

Algebra ->  Test -> SOLUTION: Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x)=12x^(2/3) -x on [0,1728]      Log On


   

Question 1155624: Determine the location and value of the absolute extreme values of f on the given
interval, if they exist. f(x)=12x^(2/3) -x on [0,1728]
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the location and value of the absolute extreme values of f on the given
interval, if they exist. f(x)=12x^(2/3) -x on [0,1728]
f%28x%29=12x%5E%282%2F3%29+-x on [0,1728] 



To find potential extrema set f'(x) = 0 



2nd derivative test for max or min

matrix%282%2C1%2C%22%22%2C%22f%27%28x%29%22=8x%5E%28-1%2F3%29-1%29





That's negative so relative maximum at x=512.

We find the value there.

f%28512%29=12%2A512%5E%282%2F3%29+-512=256

So relative maximum is (512,256)

We must examine to see if endpoints of interval are higher points.

f%280%29=12%2A0%5E%282%2F3%29+-0=0
f%281728%29=12%2A1728%5E%282%2F3%29+-1728=0

Thus absolute maximum at (512,256), absolute minima at (0,0) and (1728,0)

Edwin