We want to determine the points of inflection for a function given that its derivative is `g'(x)=(3x^2+15x+6)/((x-1)^(1/3))` and that the function's domain is `RR` .

To find the points of inflection we find the second derivative and note where the concavity of the graph changes -- i.e. when the sign of the second derivative changes.

`g''(x)=((x-1)^(1/3)(6x+15)-(3x^2+15x+6)(1/3)(x-1)^(-2/3))/((x-1)^(2/3))`

`=((x-1)^(-2/3)[(x-1)(6x+15)-(3x^2+15x+6)(1/3)])/((x-1)^(2/3))`

`=(5x^2+4x-17)/((x-1)^(4/3))`

Now `g''(x)` fails to exist at x=1, and is zero at `x=(-4+-sqrt(356))/10` or `x=-2/5+-sqrt(89)/5 ~~-2.287,1.487`

So we check the sign of the second derivative on the following intervals:

`(-oo,-2/5-sqrt(89)/5):` the second derivative is positive so the function is concave up on this interval.

`(-2/5-sqrt(89))/5,1):` the second derivative is negative so the function is concave down on this interval.

`(1,-2/5+sqrt(89)/5):` the second derivative is negative so the function is concave down on this interval.

`(-2/5+sqrt(89)/5,oo):` the second derivative is positive so the function is concave up on this interval.

**The sign of the second derivative changes at `x=-2/5-sqrt(89)/5` and at `x=-2/5+sqrt(89)/5` so these are the inflection points.**

The graph of a possible g(x) (the function could be shifted vertically without changing points of inflection, extrema, etc...):

Note that the first derivative fails at x=1 as there is a cusp there. Also, the shift from concave down to concave up at `x~~1.5` is very difficult to see from the graph.

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