Extrema and the First and Second Derivative Tests

Extreme Points

In everyday life, we often look for the "best" value, like the biggest profit, the smallest cost, or the shortest distance. In mathematics, these optimal values are known as extreme values, which consist of maximum values and minimum values. The points where these values occur are called extreme points.

An interior local extremum of a differentiable function is a stationary point, where $f'(x)=0$. More generally, local extrema may also occur where the derivative is undefined or at an endpoint of the domain. Every stationary point is a candidate for a local extremum, but it still needs to be checked. The checks below use the First and Second Derivative Tests.

First Derivative Test

This method focuses on the change in the function's behavior, i.e., whether it's increasing or decreasing, around a stationary point. By observing the sign of $f'(x)$, we can identify the type of stationary point.

• Maximum Turning Point: Occurs if the function changes from increasing to decreasing. This means the sign of $f'(x)$ changes from positive (+) to negative (-), as on the slope toward a hilltop.

• Minimum Turning Point: Occurs if the function changes from decreasing to increasing. The sign of $f'(x)$ changes from negative (-) to positive (+), as at the bottom of a valley.

• No Local Extremum: If the sign of $f'(x)$ does not change, the stationary point is neither a local maximum nor a local minimum. It may be a horizontal inflection point, but that requires a separate check that the concavity changes.

The visualization below shows the horizontal tangents at a local maximum and a local minimum.

Determine the stationary values of $f(x) = x(x-3)^2$ with the first derivative test.

Solution:

Step 1: Find the first derivative

First, expand the function: $f(x) = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x$.

Its derivative is:

Step 2: Find the stationary points

Set $f'(x) = 0$.

The stationary points occur at $x = 1$ and $x = 3$.

Step 3: Test the sign around the stationary points

• Around $x = 1$: For $x < 1$ (e.g., $x=0$), $f'(0) = 9$ (positive). For $x > 1$ (e.g., $x=2$), $f'(2) = -3$ (negative). Since the sign changes from (+) to (-), $x=1$ is a maximum turning point.

• Around $x = 3$: For $x < 3$ (e.g., $x=2$), $f'(2) = -3$ (negative). For $x > 3$ (e.g., $x=4$), $f'(4) = 9$ (positive). Since the sign changes from (-) to (+), $x=3$ is a minimum turning point.

Step 4: Determine the stationary values

To get the turning point values ($y$-values), substitute the stationary point $x$-coordinates ($x = 1$ and $x = 3$) back into the original function $f(x)$.

• Local maximum value: $f(1) = 1(1-3)^2 = 4$.

• Local minimum value: $f(3) = 3(3-3)^2 = 0$.

So, the local maximum value is $4$ at $(1,4)$, and the local minimum value is $0$ at $(3,0)$.

Second Derivative Test

This method is often faster because it doesn't require testing intervals. It uses the second derivative, $f''(x)$, to determine the concavity of the curve at a stationary point.

Suppose $f'(c) = 0$ and $f''$ is continuous on an interval containing $c$.

• If $f''(c) < 0$, then $f(c)$ is a local maximum value. The curve is concave down at that point, like an upside-down bowl.

• If $f''(c) > 0$, then $f(c)$ is a local minimum value. The curve is concave up at that point, like an open bowl.

• If $f''(c) = 0$, the test is inconclusive. Use the First Derivative Test or another argument. The point could be a maximum, a minimum, an inflection point, or none of these.

Determine the extreme values of $f(x) = \frac{1}{3}x^3 - x^2 - 3x + 2$ using the second derivative test.

Solution:

Step 1: Find the first and second derivatives

Step 2: Find the stationary points

The stationary points occur at $x = 3$ and $x = -1$.

Step 3: Test the stationary points with the second derivative

• For $x = 3$: $f''(3) = 2(3) - 2 = 4$. Since $4 > 0$, this gives a local minimum.

• For $x = -1$: $f''(-1) = 2(-1) - 2 = -4$. Since $-4 < 0$, this gives a local maximum.

Step 4: Calculate the extreme values

Substitute $x=3$ and $x=-1$ into the original function to get their extreme values.

• Local minimum value: $f(3) = \frac{1}{3}(3)^3 - (3)^2 - 3(3) + 2 = 9 - 9 - 9 + 2 = -7$.

• Local maximum value: $f(-1) = \frac{1}{3}(-1)^3 - (-1)^2 - 3(-1) + 2 = -\frac{1}{3} - 1 + 3 + 2 = \frac{11}{3}$.

The curve is concave down at the point $x=-1$ and concave up at the point $x=3$.

Exercises

Each problem asks for a stationary point or an extreme value. Set the first derivative to zero, then decide whether the point is a maximum or a minimum.

1. Determine the stationary point for the function $y = x^2 + 2x - 3$.

Worked Solution

1. Solution:

   Step 1: Find the first derivative

   $$f'(x) = 2x + 2$$

   Step 2: Find the stationary point

   Set $f'(x) = 0$.

   $$2x + 2 = 0 \implies x = -1$$

   Step 3: Find the $y$-value

   Substitute $x = -1$ into the original function.

   $$\begin{aligned} y &= (-1)^2 + 2(-1) - 3 \\ &= 1 - 2 - 3 \\ &= -4 \end{aligned}$$

   So, the stationary point is $(-1, -4)$. With the Second Derivative Test ($f''(x)=2 > 0$), this point is a minimum.