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 $f^{\prime}(x) \lt 0$ 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.

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

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

Point of Inflection (Inflection Point)

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 $f(x)=2x^3-18x^2+30x+1$.

\( \begin{align} \displaystyle
f^{\prime}(x) &= 6x^2 – 36x+30 \\
f^{\prime \prime}(x) &= 12x-36 \\
12x-36 &= 0 \\
x-3 &= 0 \\
x &= 3 \\
f^{\prime}(3) &= 6 \times 3^2 – 36 \times 3 + 30 \\
&= -24 \\
f(3) &= 2 \times 3^3 – 18 \times 3^2 + 30 \times 3+1 \\
&= -44 \\
f^{\prime}(3) &\ne 0 \text{ and } f^{\prime \prime}(3) = 0 \\
\end{align} \)
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.

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


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