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

Question

Elijah Abram · Accepted Answer

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.

Emma Aldridge · Answer

undefined To find the intervals of increasing and decreasing for the function $ y = (2x - 8)^{\frac{2}{3}} $, you first need to find its derivative. Then, analyze the sign of the derivative to determine where the function is increasing or decreasing. Taking the derivative of $ y $ with respect to $ x $, we get: $$ \frac{dy}{dx} = \frac{2}{3}(2x - 8)^{-\frac{1}{3}} \times 2 $$ Simplify the derivative: $$ \frac{dy}{dx} = \frac{4}{3}(2x - 8)^{-\frac{1}{3}} $$ Now, set the derivative equal to zero to find critical points: $$ \frac{4}{3}(2x - 8)^{-\frac{1}{3}} = 0 $$ This implies $ (2x - 8)^{-\frac{1}{3}} = 0 $, but since a real number raised to any power cannot be zero, there are no solutions. Next, analyze the sign of the derivative in different intervals to determine where the function is increasing or decreasing. Since the derivative is positive when $ (2x - 8)^{-\frac{1}{3}} > 0 $, and negative when $ (2x - 8)^{-\frac{1}{3}} < 0 $, we can conclude: 1. $ (2x - 8)^{-\frac{1}{3}} > 0 $ when $ 2x - 8 > 0 $, which gives $ x > 4 $. 2. $ (2x - 8)^{-\frac{1}{3}} < 0 $ when $ 2x - 8 < 0 $, which gives $ x < 4 $. So, the function is increasing for $ x > 4 $ and decreasing for $ x < 4 $. Therefore, the intervals of increasing and decreasing are: - Increasing: $ x > 4 $ - Decreasing: $ x < 4 $