Question

How do you find points of inflection and determine the intervals of concavity given `y=(x+2)^(1/3)`?

Answer 1

See below.

Explanation:

Finding the first derivative and solving for `0` will identify any stationary points. These can be max/min of points of inflection.

`d/dx(x+2)^1/3=1/(x+2)^(2/3)`

`f'(x)=0=> 1/(x+2)^(2/3)=0`

`0=1`

This has no solutions. ( no max/min, but possible inflection point )

Using second derivative.

`f''(x) > 0` convex ( concave up )

`f''(x) < 0` concave (concave down )

Second derivative:

`f''(x)=f'(f'x)`

`d/dx(1/(x+2)^(2/3))-2/(9(x+2)^(5/3))`

`-2/(9(x+2)^(5/3))<0`

`9(x+2)^(5/3)>0`

`x+2>0`

`x> -2`

Concave down `color(blue)([ -2 , oo ))`

`-2/(9(x+2)^(5/3))>0`

`1/((x+2)^(5/3))<0`

`(x+2)<0`

`x<-2`

Convex `color(blue)((-oo,-2])`

Inflection point at `color(blue)(((-2,0)))`

GRAPH:

graph{y=(x+2)^(1/3) [-12.66, 12.65, -6.33, 6.33]}