How do you find the inflection points of the graph of the function: # f(x)=x^(1/3)#?

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Answer 1 · 2015-08-31T09:19:43

Examine the behaviour of #f''(x) = -2/9x^(-5/3)# to find the inflexion point at #x=0#

Given #f(x) = x^(1/3)#

#f'(x) = (1/3)x^(-2/3)# which is undefined at #x=0#

#f''(x) = -2/9x^(-5/3)# which is also undefined at #x=0#

Note that, where it is defined #f''(x)# is always non-zero.

There are therefore no inflexion points in the domain of #f''(x)#, that is in #(-oo, 0) uu (0, oo)#.

However, #x=0# is an inflexion point of #f(x)#, since for any #epsilon > 0# we find:

#f''(epsilon) < 0# and #f''(-epsilon) > 0#

So the sign of #f''(x)# changes at #x = 0#