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

Question

3507 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-07T17:47:07

#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.