How do you find the critical numbers for #f(x) = x^(1/3)*(x+3)^(2/3)# to determine the maximum and minimum?

Question

How do you find the critical numbers for #f(x) =  x^(1/3)*(x+3)^(2/3)# to determine the maximum and minimum?

Christopher Agresta · Accepted Answer

Please see below.

#c# is a critical number for #f# if and only if #c# is in the domain of #f# and either #f'(c)=0# or #f'(c)# does not exist. For #f(x) = x^(1/3)(x+3)^(2/3)#, we have Domain of #f# is #(-oo,oo)# and #f'(x) = 1/3x^(-2/3)(x+3)^(2/3) + x^(1/3) 2/3(x+3)^(-1/3)# # = (x+3)^(2/3)/(3x^(2/3)) + (2x^(1/3))/(3(x+3)^(1/3))# Get a common denominator and combine to make one quotient. # = ((x+3)+2x)/(3x^(2/3)(x+3)^(1/3))# # = (x+1)/(x^(2/3)(x+3)^(1/3))# #f'(x) = (x+1)/(x^(2/3)(x+3)^(1/3))# is #0# at #x=-1# and #f'(x)# does not exist for #x=-3#, and #x=0# These are all in the domain of #f#, so the critical numbers are: #x=-3# (local max) #x=-1# (local min) #x=0# (neither min nor max) The graph of #f# is shown below. graph{ x^(1/3)(x+3)^(2/3) [-7.024, 7.02, -3.51, 3.514]}

Christopher Agresta · Answer

undefined To find the critical numbers for $ f(x) = x^{1/3} \cdot (x+3)^{2/3} $ and determine the maximum and minimum, follow these steps:

1. Compute the derivative of $ f(x) $ using the product rule and simplify the expression.
2. Set the derivative equal to zero and solve for $ x $ to find the critical numbers.
3. Evaluate $ f(x) $ at each critical number and at the endpoints of the domain to determine the maximum and minimum values of the function.