Question

What are the points of inflection, if any, of `f(x) = x^4/12 - 2x^2 + 15`?

Answer 1

The function `f(x) = x^4/12-2x^2+15` has two inflection points in `x_1=-2` and `x_2=2`.

Explanation:

A necessary condition for the curve `y=f(x)` to have an inflection point for `x=barx` is that:

`f''(barx) = 0`

Evaluate the second derivative of the function:

`f(x) = x^4/12-2x^2+15`

`f'(x) = x^3/3-4x`

`f''(x) = x^2 -4`

From the equation:

`x^2 - 4 = 0`

`x^2 = 4`

`x=+-2`

we get that the function may have an inflection point in `x_1 = -2` and `x_2 = 2`.

Now consider the inequality:

`f''(x) > 0`

As `f''(x)` is a second degree polynomial with positive leading coefficient, we have that its value is negative in the interval between the roots and positive outside.

`f''(x)` then changes sign both around `x_1` and `x_2`, so these are actually inflection points.

Answer 2

There are two non-stationary points of inflection which occur at `(+-2, 25/3)`

Explanation:

We have:

`f(x) = x^4/12-2x^2+15`

We would normally look for critical points, that is coordinates where `f'(x)=0`, however the question does not require this so it will be skipped.

The first derivative is then:

`f'(x) = x^3/3-4x`

So the second derivative is then:

`f''(x) = x^2-4`

We look for inflection points , which are coordinates where the second derivative vanishes:

`f''(x) = 0 => x^2-4 = 0`
`:. x^2=4`
`:. x = +-2`

A point of inflection is graded as a stationary point of inflection of the first derivative vanished at the point, otherwise as a non-stationary point of inflection

So when `x=-2`, we have:

`f(-2) = 16/12-8+15 = 25/3`
`f'(-2) = -8/3+8 = 16/3`

And when `x=2`:

`f'(2) = 16/12-8+15=25/3`
`f'(2) = 8/3-8 = 16/3`

Hence, There are two non-stationary points of inflection which occur at `(+-2, 25/3)`

It can be interesting to see the graphs of the function compared with the first and second derivatives:

The graph of the function `y=f(x)`

graph{x^4/12-2x^2+15 [-6, 6, -10, 18]}

The graph of the function `y=f'(x)`. The zeros coincide with critical points of the curve `y=f(x)` ie the turning points of `y=f(x)`

graph{x^3/3-4x [-6, 6, -10, 18]}

The graph of the function `y=f''(x)`. The zeros coincide with critical points of the curve `y=f'(x)` ie the turning points of `y=f'(x)` or the points of inflection of `y=f(x)`

graph{x^2-4 [-6, 6, -10, 18]}