A function’s turning point is where $f'(x)=0$.

A maximum turning point is a turning point where the curve is concave up (from increasing to decreasing ) and $f^{\prime}(x)=0$ at the point.
$$ \begin{array}{|c|c|c|} \hline
f^{\prime}(x) \gt 0 & f'(x) = 0 & f'(x) \lt 0 \\ \hline
& \text{maximum} & \\
\nearrow & & \searrow \\ \hline
\end{array} $$
A minimum turning point is a turning point where the curve is concave down (from decreasing to increasing) and $f^{\prime}(x)=0$ at the point.
$$ \begin{array}{|c|c|c|} \hline
f^{\prime}(x) \lt 0 & f^{\prime}(x) = 0 & f^{\prime}(x) \gt 0 \\ \hline
\searrow & & \nearrow \\
& \text{minimum} & \\ \hline
\end{array} $$

Example 1

Find any turning points and their nature of $f(x) =2x^3-9x^2+12x+3$.

\( \begin{align} \displaystyle \require{color}
f^{\prime}(x) &= 6x^2-18x+12 \\
6x^2-18x+12 &= 0 \\
x^2-3x+2 &= 0 \\
(x-1)(x-2) &= 0 \\
x &= 1 \text{ or } x=2 \\
f(1) &= 2 \times 1^3-9 \times 1^2+12 \times 1+3 \\
&= 8 \\
f(2) &= 2 \times 2^3-9 \times 2^2+12 \times 2+3 \\
&= 7 \\
\end{align} \)
There are two turning points; $(1,8)$ and $(2,7)$.
\( \begin{align} \displaystyle \require{color}
f^{\prime}(0) &= 6 \times 0^2-18 \times 0 + 12 \\
&= +12 \\
f^{\prime}(1.5) &= 6 \times 1.5^2-18 \times 1.5 + 12 \\
&= -1.5 \\
f^{\prime}(3) &= 6 \times 3^2-18 \times 3 + 12 \\
&= +12
\end{align} \)
$$ \begin{array}{|c|c|c|c|c|c|} \hline
x & 0 & 1 & 1.5 & 2 & 3 \\ \hline
f^{\prime}(x) & +12 & 0 & -1.5 & 0 & +12 \\ \hline
\text{shape} & \nearrow & \text{max} & \searrow & \text{min} & \nearrow \\ \hline
\end{array} $$
Therefore $(1,8)$ is a maximum turning point and $(2,7)$ is a minimum turning point.

Example 2

Find any turning points and their nature of $f(x) =-x^3+6x^2-9x-5$.

\( \begin{align} \displaystyle
f^{\prime}(x) &= -3x^2+12x-9 \\
-3x^2+12x-9 &= 0 \\
x^2+4x-3 &= 0 \\
(x-1)(x-3) &= 0 \\
x &= 1 \text{ or } x=3 \\
f^{\prime}(0) &= -3 \times 0^2 + 12 \times 0-9 \\
&= -9 \\
f^{\prime}(2) &= -3 \times 2^2 + 12 \times 2-9 \\
&= +3 \\
f^{\prime}(4) &= -3 \times 4^2 + 12 \times 4-9 \\
&= -9
\end{align} \)
$$ \begin{array}{|c|c|c|c|c|c|} \hline
x & 0 & 1 & 2 & 3 & 4 \\ \hline
f^{\prime}(x) & -9 & 0 & +3 & 0 & -9 \\ \hline
\text{shape} & \searrow & \text{min} & \nearrow & \text{max} & \searrow \\ \hline
\end{array} $$
\( \begin{align}
f(1) & = -1^3 +6 \times 1^2-9 \times 1-5 \\
&= -9 \\
f(3) & = -3^3 +6 \times 3^2-9 \times 3-5 \\
&= -5
\end{align} \)
Therefore $(1,-9)$ is a minimum turning point and $(3,-5)$ is a maximum turning point.

Determining the Nature of Turning Points by Concavities

\begin{array}{|c|c|} \hline
\text{concave downwards} & \text{concave upwards} \\ \hline
\cup \text{ shape} & \cap \text{ shape} \\ \hline
f^{\prime \prime}(x) \gt 0 & f^{\prime \prime}(x) \lt 0 \\ \hline
\end{array}
A maximum turning point is a turning point where the curve is concave upwards, $f^{\prime \prime}(x) \lt 0$ and $f^{\prime}(x)=0$ at the point.
A minimum turning point is a turning point where the curve is concave downwards, $f^{\prime \prime}(x) \gt 0$ and $f^{\prime}(x)=0$ at the point.

$$ \begin{array}{|c|c|} \hline
\text{maximum turning point} & \text{minimum turning point} \\ \hline
f^{\prime}(x) = 0 & f'(x) = 0 \\ \hline
f^{\prime \prime}(x) \lt 0 & f^{\prime \prime}(x) \gt 0 \\ \hline
\end{array} $$

Example 3

Use second derivatives to find any turning points and their nature of $f(x) =2x^3-9x^2+12x+3$.

\( \begin{align} \displaystyle \require{color}
f^{\prime}(x) &= 6x^2-18x+12 \\
6x^2-18x+12 &= 0 \\
x^2-3x+2 &= 0 \\
(x-1)(x-2) &= 0 \\
x &= 1 \text{ or } x=2 \\
f(1) &= 2 \times 1^3-9 \times 1^2+12 \times 1+3 \\
&= 8 \\
f(2) &= 2 \times 2^3-9 \times 2^2+12 \times 2+3 \\
&= 7
\end{align} \)
There are two turning points; $(1,8)$ and $(2,7)$.
\( \begin{align} \displaystyle \require{color}
f^{\prime \prime}(x) &= 12x-18 \\
f^{\prime \prime}(1) &= 12 \times 1-18 \\
&= -6 \lt 0 \\
\end{align} \)
This indicates that point $(1,8)$ is a maximum turning point.
\( \begin{align} \displaystyle \require{color}
f^{\prime \prime}(2) &= 12 \times 2-18 \\
&= +6 \gt 0
\end{align} \)
This indicates that point $(2,7)$ is a minimum turning point.

Example 4

Use second derivatives to find any turning points and their nature of $f(x) =-x^3+6x^2-9x-5$.

\( \begin{align} \displaystyle
f^{\prime}(x) &= -3x^2+12x-9 \\
-3x^2+12x-9 &= 0 \\
x^2+4x-3 &= 0 \\
(x-1)(x-3) &= 0 \\
x &= 1 \text{ or } x=3 \\
f(1) & = -1^3 +6 \times 1^2-9 \times 1-5 \\
&= -9 \\
f(3) & = -3^3 +6 \times 3^2-9 \times 3-5 \\
&= -5
\end{align} \)
There are two turning points; $(1,-9)$ and $(3,-5)$.
\( \begin{align} \displaystyle
f^{\prime \prime}(x) &= -6x+12 \\
f^{\prime \prime}(1) &= -6 \times 1 + 12 \\
&= +6 \gt 0 \\
\end{align} \)
This indicates that point $(1,-9)$ is a minimum turning point.
\( \begin{align} \displaystyle
f^{\prime \prime}(3) &= -6 \times 3 + 12 \\
&= -6 \lt 0
\end{align} \)
This indicates that point $(3,-5)$ is a maximum turning point.

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