# 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.

The gradient is positive on either side of the stationary point.

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

The gradient is negative on 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 on either side of any stationary points. This is known as the first derivative test.

### Example 1

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

\begin{align} \displaystyle f^{\prime}(x) &= 6x^2-36x+30 \\ f^{\prime \prime}(x) &= 12x-36 \\ 12x-36 &= 0 \\ x-3 &= 0 \\ x &= 3 \\ f^{\prime}(3) &= 6 \times 3^2-36 \times 3 + 30 \\ &= -24 \\ f(3) &= 2 \times 3^3-18 \times 3^2 + 30 \times 3+1 \\ &= -44 \\ f^{\prime}(3) &\ne 0 \text{ and } f^{\prime \prime}(3) = 0 \end{align}
Therefore the point $(3,-44)$ is a non-horizontal point of inflection.

### Example 2

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

\begin{align} \displaystyle f^{\prime}(x) &= 3x^2+12x+12 \\ f^{\prime \prime}(x) &= 6x+12 \\ 6x+12 &= 0 \\ x+2 &= 0 \\ x &= -2 \\ f^{\prime}(-2) &= 3 \times (-2)^2 + 12 \times (-2) + 12 \\ &= 0 \\ f(-2) &= (-2)^3 + 6 \times (-2)^2 + 12 \times (-2) + 7 \\ &= -1 \\ f^{\prime}(-2) &= 0 \text{ and } f^{\prime \prime}(-2) = 0 \end{align}
Therefore the point $(-2,-1)$ is a horizontal point of inflection.

### Example 3

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

\begin{align} \displaystyle f^{\prime}(x) &= 4x^3 + 6x^2-24x-2 \\ f^{\prime \prime}(x) &= 12x^2 + 12x-24 \\ 12x^2 + 12x-24 &\gt 0 &\text{concave upwards} \\ x^2 + x-2 &\gt 0 \\ (x+2)(x-1) &\gt 0 \\ x &\lt -2 \text{ or } x \gt 1 &\text{concave upwards} \\ -2 &\lt x \lt 1 &\text{concave downwards} \end{align}