Let f(x)= 8(x-8)^(2/3) + 8. find the open intervals on which f is increasing (decreasing). then determine the

let f(x)= 8(x-8)^(2/3) + 8. find the open intervals on which f is increasing (decreasing). then determine the x-coordinates of all relative maxima (minima).
f is increasing intervals......
f is decreasing intervals......

Take the derivative:
16/(3(x-8)^(1/3)) = d/dx

Then set equal to zero to find the max and mins
16/(3(x-8)^(1/3)) = 0

Not equal to zero anywhere, move on to critical points (discontinuity)
at x=8 there is an infinite discontinuity (in this graph it is technically a cusp)
16/(3(8-8)^(1/3)) = 16/0

Now we check if it is positive or negative slope to the left and right.

x=7
16/(3(7-8)^(1/3)) = 16/(-3) = negative number. So the slope on the left is negative

x=9
16/(3(9-8)^(1/3)) = 16/3 = positive number. So the slope is positive on the right

Since this is a cusp, and the left is negative and the right is positive
relative minima at x=8
negative: (-infinity,8)
positive: (8,infinity)