How do you find the intervals of increasing and decreasing given #y=(2x-8)^(2/3)#?

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Answer 1 · 2017-08-07T14:29:31

Please see below.

#y=(2x-8)^(2/3)#

The domain is #(-oo,oo)#.

#y' = 2/3(2x-8)^(-1/3) * d/dx(2x-8)#

# = 4/3(2x-8)^(-1/3)#

# = 4/(3root(3)(2x-8)#

#y'# is never #0# and

#y'# is undefined (does not exist) at #x = 4#

So we look at the sign of #y'# on both sides of #4#

On #(-oo,4)#, #y' < 0# so the function is decreasing.

On #(4,oo)#, #y' > 0# so the function is increasing.

Bonus

From the above analysis we see that #f(4) = 0# is a local minimum. In fact it is a global minimum as well.