# Inflection Points (Points of Inflection)

## Horizontal (stationary) point of inflection (inflection point)

If $x \lt a$, then $f'(x) \gt 0$ and $f^{\prime \prime}(x) \le 0 \rightarrow$ concave down.
If $x = a$, then $f'(x) = 0$ and $f^{\prime \prime}(x) = 0 \rightarrow$ horizontal point inflection.
If $x \gt a$, then $f'(x) \gt 0$ and $f^{\prime \prime}(x) \ge 0 \rightarrow$ concave up.

That is, the gradient is positive either side of the stationary point.

If $x \lt a$, then $f'(x) \lt 0$ and $f^{\prime \prime}(x) \ge 0 \rightarrow$ concave up.
If $x = a$, then $f'(x) = 0$ and $f^{\prime \prime}(x)=0 \rightarrow$ horizontal point of inflection.
If $x \gt a$, then $f'(x) \lt 0$ and $f^{\prime \prime}(x) \le 0 \rightarrow$ concave down.

That is, the gradient is negative either side of the stationary point.

## Not all points of inflection (inflection points) are stationary points

The gradient of the tangent is not equal to 0. At the point of inflection, $f'(x) \ne 0$ and $f^{\prime \prime}(x)=0$.
When determining the nature of stationary points it is helpful to complete a ‘gradient table’, which shows the sign of the gradient either side of any stationary points. This is known as the first derivative test.

### Example 1

Find the points of inflection (inflection points) of $f(x)=2x^3-18x^2+30x+1$.

### Example 2

Find the points of inflection (inflection points) of $f(x)=x^3+6x^2+12x+7$.

### Example 3

Find intervals where $f(x)=x^4 + 2x^3 – 12x^2 – 2x + 7$ is concave up or down.