Can an inflection point occur where $f''(x)$ is undefined?

Yes, if the function is continuous at that point and the concavity changes. An example is $f(x) = x^{1/3}$ at $x=0$.

What if $f''(x) = 0$ but the sign doesn't change?

The point is not an inflection point. The curve maintains the same concavity on both sides of the point.

Do asymptotes count as inflection points?

No. Even if the concavity changes across a vertical asymptote, the function is not continuous there, so there is no point on the curve to label as an inflection point.

How to find inflection points

An inflection point occurs where a curve changes concavity, which requires the second derivative to change sign. This method applies to any twice-differentiable function on a given continuous interval.

The setup

You need the function $f(x)$ and its domain. Verify that $f(x)$ is continuous across the domain where you are testing for inflection points.

The steps

Compute the first derivative $f'(x)$ .
Compute the second derivative $f''(x)$ .
Find all candidate points by setting $f''(x) = 0$ or identifying where $f''(x)$ is undefined.
Set up a sign chart for $f''(x)$ using test points in the intervals between the candidate points.
If $f''(x)$ changes sign at $x=c$ (from positive to negative, or negative to positive) and $f(c)$ exists, then $(c, f(c))$ is an inflection point.

Checking the result

Evaluate $f''(x)$ at test points strictly on either side of the candidate $x=c$ . The sign of $f''(x)$ must be definitively positive on one side and negative on the other. Verify $f(c)$ is a real number.

Common errors

Assuming $f''(c) = 0$ guarantees an inflection point; it does not if the sign of $f''(x)$ does not change (e.g., $f(x) = x^4$ ). Another error is forgetting to check if the original function $f(x)$ is defined at the candidate point $x=c$ , such as at vertical asymptotes.

Worked example

Find the inflection points of $f(x) = x^4 - 4x^3$ .

Compute the first derivative: $f'(x) = 4x^3 - 12x^2$

Compute the second derivative: $f''(x) = 12x^2 - 24x$

Set $f''(x) = 0$ to find candidate points: $12x(x - 2) = 0$ $x = 0$ and $x = 2$

Test intervals around candidates: For $x < 0$ , let $x = -1$ : $f''(-1) = 12(-1)^2 - 24(-1) = 36 > 0$ . For $0 < x < 2$ , let $x = 1$ : $f''(1) = 12(1)^2 - 24(1) = -12 < 0$ . For $x > 2$ , let $x = 3$ : $f''(3) = 12(3)^2 - 24(3) = 36 > 0$ .

The second derivative changes sign at $x = 0$ and $x = 2$ .

Find the $y$ -coordinates: $f(0) = 0^4 - 4(0)^3 = 0$ $f(2) = 2^4 - 4(2)^3 = 16 - 32 = -16$

The inflection points are $(0, 0)$ and $(2, -16)$ .

FAQ

Run your own problem

Answer form

References: Calculus: Early Transcendentals by James Stewart · OpenStax Calculus Volume 1 · Khan Academy: Applications of derivatives