Question

How do you find the inflection points of the graph of the function: `y= (1/x^2) - (1/x^3)`?

Answer 1

Find the points on the curve at which the concavity changes.

The only inflection point is `(2, 1/8)`

Explanation:

The terminology used in calculus classes at schools I've worked at in the US (where the question was asked) is that an inflection point is a point on the graph at which the concavity changes.

Let `f(x) = y = 1/x^2 - 1/x^3` (so it is clear what I mean when i say "the function")

This is not a nice form for investigating concavity using the second derivative, so rewrite the function in a nicer form:

`f(x) = (x-1)/x^3`

We need to investigate the sign of `f''(x)`, so

`f'(x) = ((1)(x^3) - 3x^2(x-1))/(x^3)^2 = (-2x^3+3x^2)/x^6 = (-2x+3)/x^4`

and

`f''(x) = ((-2)(x^4)-(4x^3)(-2x+3))/(x^4)^2 = (6x^4-12x^3)/x^8 = (6x-12)/x^5`

The sign of `f''(x)` changes at `x=0` and at `x=2`.

There is no point on the graph of this function when `x=0`, so there is no inflection point with `x`-coordinate `0`.

At `x=2`, we get `y = f(2) = 1/8`, so `(2, 1/8)` is the only inflection point.