Question 996110
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
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If x = 1, then
f ' (x) = (80/3)*x^(5/3)
f ' (1) = (80/3)*(1)^(5/3)
f ' (1) = 26.667 ... positive result
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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)