SOLUTION: Let f (x) = 10x^(8/3) Find all extreme values (if any) of f on the interval [-8, 8]. Determine at which numbers in the interval these values occur. NOTE: If there is no maximum

Question 996110: Let f (x) = 10x^(8/3)
Find all extreme values (if any) of f on the interval [-8, 8]. Determine at which numbers in the interval these values occur. NOTE: If there is no maximum value (minimum value), then enter N/A in the corresponding blanks.
Maximum Value:
f (?)= ?= f (?)
Minimum Value:
f (?) = ?

Please explain
Thank you
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Apply the derivative, then solve f ' (x) = 0
f(x) = 10x^(8/3)
f ' (x) = 10*(8/3)*x^(8/3-1)
f ' (x) = (80/3)*x^(5/3)
0 = (80/3)*x^(5/3)
(80/3)*x^(5/3) = 0
x^(5/3) = 0
[x^(5/3)]^(3/5) = 0^(3/5)
x = 0^(3/5)
x = 0
The only critical value is x = 0

Make a sign chart for f ' (x) with the critical value 0 plotted on the number line. Also plot -1 and +1 (one number on each side of the critical value)
If x = -1, then
f ' (x) = (80/3)*x^(5/3)
f ' (-1) = (80/3)*(-1)^(5/3)
f ' (-1) = -26.667 ... negative result
---------------
If x = 1, then
f ' (x) = (80/3)*x^(5/3)
f ' (1) = (80/3)*(1)^(5/3)
f ' (1) = 26.667 ... positive result
---------------
We have f ' (x) changing sign as it passes through x = 0
f ' (x) changes from negative to positive, so we have a local min when x = 0

f(x) = 10x^(8/3)
f(0) = 10*0^(8/3)
f(0) = 0

The local min is at the point (0,0)
There is NO local max since there is only one extrema (from the one critical value)
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