Finding Extrema

Section 3.3 Finding Extrema

Standards

Find extrema using derivatives

Section 3.1 and Section 3.2 introduce connections between derivatives and their functions including possible locations of relative extrema. In this section examples of using these traits to find and indentify the types.

Subsection 3.3.1 Finding Relative Extrema

We will find the relative extrema for

\begin{equation*} \omega(t) = \sqrt[3]{(x-1)^2(x-3)^2} \end{equation*}

using three different methods.

For any method for finding relative (local) extrema we need to determine where the derivative is zero (horizontal tangents) and undefined (cusps).

\begin{align*} \omega(t) = \amp \sqrt[3]{(t-1)^2(t-3)^2}\\ = \amp ((t-1)^2(t-3)^2)^{1/3}. \\ \omega^\prime(t) = \amp \frac{1}{3}((t-1)^2(t-3)^2)^{-2/3} \cdot [(t-1)^2 \cdot 2(t-3) \cdot 1+(t-3)^2 \cdot 2(t-1) \cdot 1]\\ = \amp \frac{(t-1)^2(t-3)+(t-3)^2(t-1)}{3((t-1)^2(t-3)^2)^{2/3}}\\ = \amp \frac{(t-1)(t-3)[(t-1)+(t-3)]}{3((t-1)^2(t-3)^2)^{2/3}}\\ = \amp \frac{(t-1)(t-3)(2t-4)}{3((t-1)^2(t-3)^2)^{2/3}}\\ = \amp \frac{2(t-1)(t-3)(t-2)}{3((t-1)^2(t-3)^2)^{2/3}}. \end{align*}

The derivative will be zero only where the numerator is zero which occurs at \(x=1,2,3\text{.}\) The derivative will be undefined where the denominator is zero which occur at \(x=1,3\text{.}\) Thus possible locations for relative extrema are \(x=1,2,3\text{.}\)

Example 3.3.1. Identifying Relative Extrema via Function Values.

To determine which are relative maxima, which are relative minima, and which are something else we need to test points.

Solution.

After using the derivative to determine where relative extrema might occur we can use values of the function. For this test we evaluate the function at the critical values and left and right of the critical values.

Table 3.3.2. Functions Values to Determine Extrema

\(t\)	\(\omega(t)\)
\(\frac{1}{2}\)	1.16
\(1\)	0
\(\frac{3}{2}\)	0.83
\(2\)	1
\(\frac{5}{2}\)	0.83
\(3\)	0
\(\frac{7}{2}\)	1.16

The relative extrema are as follows.

Because the points left and right of \(t=1\) are higher, \(t=1\) is the location of a relative minimum.
Because the points left and right of \(t=2\) are lower, \(t=2\) is the location of a relative maximum.
Because the points left and right of \(t=3\) are higher, \(t=3\) is the location of a relative minimum.

Example 3.3.3. Identifying Relative Extrema via 1st Derivative.

To determine which possible extrema are relative maxima, which are relative minima, and which are something else we can use the first derivative.

Solution.

After using the derivative to determine where relative extrema might occur we can use values of the derivative to determine where the function is increasing and decreasing. For this test we evaluate the derivative left and right of each critical value.

Table 3.3.4. Derivative Values to Determine Extrema

\(t\)	\(\omega^\prime(t)\)
\(\frac{1}{2}\)	-
\(1\)	undefined
\(\frac{3}{2}\)	+
\(2\)	0
\(\frac{5}{2}\)	-
\(3\)	undefined
\(\frac{7}{2}\)	+

The relative extrema are as follows.

Because the curve decreases on the left and increases on the right of \(t=1\) this is the location of a relative minimum.
Because the curve increases on the left and decreases on the right of \(t=2\) this is the location of a relative maximum.
Because the curve decreases on the left and increases on the right of \(t=3\) this is the location of a relative minimum.

Example 3.3.5. Identifying Relative Extrema via 2nd Derivative.

To determine which possible extrema are relative maxima, which are relative minima, and which are something else we can use the second derivative.

Solution.

After using the derivative to determine where relative extrema might occur we can use values of the second derivative to determine which are maxima and which are minima. For this test we evaluate the second derivative at the critical values found.

Table 3.3.6. Derivative Values to Determine Extrema

\(1\)	undefined
\(2\)	-
\(3\)	undefined

The second derivative only helps us at \(t=2\text{.}\) Because the curve is concave down, this is a relative maximum. The other values are undefined (cusps), so this test is not helpful.

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