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)#?

Question

Charles Anctil · Accepted Answer

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)# There is not a point of inflection because the concavity remains constant. This is the graph: [-10, 10, -5, 4.995]} graph{-(x+2)^(2/3)

Emilia Aguilar · Answer

undefined To find the x-coordinates of points of inflection, we need to find where the second derivative changes sign or is zero. Given $ y = -(x + 2)^{2/3} $, First derivative: $$ \frac{dy}{dx} = -\frac{2}{3}(x + 2)^{-1/3} $$ Second derivative: $$ \frac{d^2y}{dx^2} = \frac{2}{9}(x + 2)^{-4/3} $$ Set the second derivative to zero to find potential points of inflection: $$ \frac{2}{9}(x + 2)^{-4/3} = 0 $$ $$ (x + 2)^{-4/3} = 0 $$ $$ \text{No real solutions since a number to any power can't be zero.} $$ To find the discontinuities: The function is continuous everywhere since it is a polynomial expression. To find the intervals of concavity, set the second derivative greater than zero for concave up and less than zero for concave down: For concave up: $$ \frac{2}{9}(x + 2)^{-4/3} > 0 $$ $$ (x + 2)^{-4/3} > 0 $$ $$ \text{No real solutions since a number to any power can't be negative.} $$ For concave down: $$ \frac{2}{9}(x + 2)^{-4/3} < 0 $$ $$ (x + 2)^{-4/3} < 0 $$ $$ \text{No real solutions since a number to any power can't be negative.} $$ Given the function, there are no points of inflection, no discontinuities, and no intervals of concavity for $ y = -(x + 2)^{2/3} $.