If #f(x) = (5/2)x^(2/3) - x^(5/3)#, what are the points of inflection, concavity and critical points?

Question

If  #f(x) = (5/2)x^(2/3) - x^(5/3)#, what are the points of inflection, concavity and critical points?

Elijah Abram · Accepted Answer

#f(x) = (5/2)x^(2/3) - x^(5/3)# The domain of #f# is #(-oo,oo)#. #f'(x) = (5/3)x^(-1/3) - 5/3 x^(2/3)# # = 5/3[x^(-1/3) - x^(2/3)]# # = 5/3[(1-x)/x^(1/3)]# #f'# does not exist at #x=0# and #f'(x) = 0# at #x=1#. Both #0# and #1# are in the domain of #f#, so both are critical numbers. #f''(x) = 5/3[-1/3x^(-4/3)-2/3x^(-1/3)]# # = -5/9x^(-4/3)[1+2x]# # = -(5(1+2x))/(9x^(4/3))# #f''(x) > 0# for #x < -1/2# and #f''(x) > 0# for #-1/2 < x < 0# and #x > 0# The graph of #f# is concave up on #(-oo,-1/2)# and concave down on #(-1/2,0)# and on #(0,oo)#. The point #(-1/2,f(-1/2))# is an inflection point.

Evelyn Agard · Answer

undefined To find the points of inflection, concavity, and critical points of the function $ f(x) = \frac{5}{2}x^\frac{2}{3} - x^\frac{5}{3} $:

1. Critical Points:
   Find the first derivative $ f'(x) $ and solve for $ f'(x) = 0 $ to find critical points.
   
   $ f'(x) = \frac{5}{3}x^{-\frac{1}{3}} - \frac{5}{3}x^\frac{2}{3} $

Setting $ f'(x) = 0 $:
   $ \frac{5}{3}x^{-\frac{1}{3}} - \frac{5}{3}x^\frac{2}{3} = 0 $
   
   $ \frac{5}{3}x^{-\frac{1}{3}} = \frac{5}{3}x^\frac{2}{3} $
   
   $ x^{-\frac{1}{3}} = x^\frac{2}{3} $
   
   $ x^{-\frac{1}{3}} - x^\frac{2}{3} = 0 $
   
   Solve for $ x $ to find critical points.

2. Concavity:
   Find the second derivative $ f''(x) $ and determine where it's positive and negative to identify intervals of concavity.
   
   $ f''(x) = -\frac{5}{9}x^{-\frac{4}{3}} - \frac{10}{9}x^{-\frac{1}{3}} $
   
   Determine the sign of $ f''(x) $ in the intervals determined by the critical points.

3. Points of Inflection:
   Points of inflection occur where the concavity changes. Find the x-values where $ f''(x) = 0 $ or where the concavity changes sign.