How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for #y=-(x+2)^(2/3)#?

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Answer 1 · 2018-01-12T17:46:23

Please see below.

Let #f(x) = -(x+2)^(2/3)# and note that #f(x) = -root(3)((x+2)^2)#.

#f# has domain: all real numbers and is continuous everywhere.

To investigate concavity look at #f''(x)#

#f'(x) = -2/3(x+2)^(-1/3)# and

#f''(x) = 2/9(x+2)^(-4/3) = 2/9 1/root(3)((x+2)^4)#.

Since #f''(x)# is positive for #x != -2# (it is undefined at #-2#), the graph of #f# is concave up on #(-oo,-2)# and also on #(-2,oo)#

Since the concavity does not change, there is no point of inflection.

Here is the graph:
graph{-(x+2)^(2/3) [-10, 10, -5, 4.995]}