# Turning Points and Nature

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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