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

Jun 17, 2015

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

The only inflection point is $\left(2 , \frac{1}{8}\right)$

#### 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 $f \left(x\right) = y = \frac{1}{x} ^ 2 - \frac{1}{x} ^ 3$ (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:

$f \left(x\right) = \frac{x - 1}{x} ^ 3$

We need to investigate the sign of $f ' ' \left(x\right)$, so

$f ' \left(x\right) = \frac{\left(1\right) \left({x}^{3}\right) - 3 {x}^{2} \left(x - 1\right)}{{x}^{3}} ^ 2 = \frac{- 2 {x}^{3} + 3 {x}^{2}}{x} ^ 6 = \frac{- 2 x + 3}{x} ^ 4$

and

$f ' ' \left(x\right) = \frac{\left(- 2\right) \left({x}^{4}\right) - \left(4 {x}^{3}\right) \left(- 2 x + 3\right)}{{x}^{4}} ^ 2 = \frac{6 {x}^{4} - 12 {x}^{3}}{x} ^ 8 = \frac{6 x - 12}{x} ^ 5$

The sign of $f ' ' \left(x\right)$ 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 $y = f \left(2\right) = \frac{1}{8}$, so $\left(2 , \frac{1}{8}\right)$ is the only inflection point.