Extrema

Section 10.1 Extrema

Subsection 10.1.1 Local Minima and Maxima

Let \(f: [a,b] \rightarrow \RR\) be a function.

A point \((c,f(c))\) is called a local maximum for \(f\) if \(f(c) \geq f(x)\) for all \(x\) close to \(c\text{.}\)
A point is a local minimum if \(f(c) \leq f(x)\) for all \(x\) close to \(c\text{.}\)

The plurals of these terms are local minima and local maxima; if I want to refer to both, I will simply say local extrema. I can also define global extrema.

The point \((c,f(c))\) is a global maximum if \(f(c) > f(x)\) for all \(x\) in the domain of the function.
The point is a global minimum if \(f(c) \lt f(x)\) for all \(x\) in the domain.

One of the most important uses of derivatives is finding local extrema. The reason that derivatives are useful is that local extrema (for differentiable functions) are identified by their tangent lines. Look at the tangent lines Figure 10.1.1.

Figure 10.1.1. Tangent Lines at Extrema

The tangent lines are horizontal at maximum or minimum values. I can make use of this fact to try to find these extrema. Horizontal tangent lines have zero slope, to they can only occur when \(f^\prime(x) = 0\text{.}\) Therefore, I look for points \(x\) where the derivative vanishes. Such points will be called the critical points of the function.

When I solve \(f^\prime(x) = 0\) to find the critical points, I don’t know if these critical points are minima, maxima or neither. However, I can look at the intervals between the critical points. I usually do this in a table of interval, demonstrated in Example 10.1.2 and Example 10.1.3. I use the critical points to split up the domain into intervals. It is important that I split the domain into intervals: if the domain already has breaks in it, these are further split up by the critical points.

Once I have these intervals, I will test the derivative on each interval. Between the critical points (and any division of the domain), the derivative will either be entirely positive or entirely negative. Therefore, I can test any point I wish in the interval. If I find that the derivative is positive, I know the function is increasing. If I find the derivative is negative, then the function is decreasing. Determining this information is called finding the intervals of increase and decrease for the function, which is itself very useful information for understanding the behaviour of the function.

The intervals tell me about the function at it critical points.

If the interval on the left of a critical is increasing and the interval on the right of a critical point is decreasing, then the critical point is a maximum. (The function changes from going up to the critical point to going down after it.)
If the interval on the left of a critical is decreasing and the interval on the right of a critical point is increasing, then the critical point is a minimum. (The function changes from going downto the critical point to going up after it.)
If the function is increasing on both sides of the critical point, or similar if it is decreasing on both sides of the critical point, the critical is neither a minimum nor a maxiumu.

A example of the last situation is the point \((0,0)\) on the graph of the cubic \(f(x) = x^3\text{.}\) There is a critical points there and the tangent line is momentarily flat, but the function is increasing on both sides and there isn’t a minimum or a maximum.

All of this gives us a nice algorithm for calculating extrema of functions, as well as finding their intervals of increase and decrease. Let me recap the algorithm for a differentiable function \(f(x)\text{.}\)

I calculate the derivative \(f^\prime(x)\)
I set the derivative equal to zero and solve for the critical points.
I use the critical point to break up the domain into intervals.
I test the derivative on each interval (using any point I want in the interval) to determine if it is positive or negative. Where is it positive, I label the interval as increasing; where is it negative, I label the interval as decreasing.
I use the intervals to determine which critical points are maxima, which are minima, and which are neither.

Subsection 10.1.2 Examples

Example 10.1.2.

Consider the quadratic \(f(x) = x^2 + 2\text{.}\) The derivative is \(f^\prime(x) = 2x\text{.}\) To find the critical points, I set the derivative equal to zero.

\begin{align*} f(x) \amp = x^2 + 2\\ f^\prime(x) \amp = 2x\\ 2x \amp = 0 \implies x = 0 \end{align*}

So \(x=0\) is the only critical point. After I have found a critical point, I need to determine if it is a maximum, minimum, or neither. I do this by looking at the sign of the derivative near the critical point. The easiest way to record this information is in intevals of increase and decrease. I divide the domain of the function into pieces separated by the critical points.

In this example, the domain is \(\RR\) and the only critical point is \(x=0\text{.}\) Splitting the domain by that critical point gives the intervals \((-\infty, 0)\) and \((0,\infty)\text{.}\) I set up a table to test the intervals.

\begin{align*} \amp (-\infty, 0) \amp \amp (0, \infty) \\ \amp f^\prime(-1) = 2(-1) = -2 \amp \amp f^\prime (1) = 2(1) = 2 \\ \amp f^\prime \lt 0 \amp \amp f^\prime \gt 0 \\ \amp \text{decreasing} \amp \amp \text{increasing} \end{align*}

This table lets me classify the critical point. At \(x=0\text{,}\) the function switches from decreasing to increasing. This means that \(x\) must be a minimum. I can conclude that \((0,2)\) is a local minimum of the function.

Example 10.1.3.

Now I’ll consider a cubic \(f(x) = 2x^3 + 3x^2 - 36x + 4\) with derivative \(f^\prime(x) = 6x^2 + 6x - 36 = 6(x^2 + x - 6)\text{.}\) I set the derivative equal to zero to find the critical points.

\begin{align*} 6 (x^2 + x - 6) \amp = 0\\ x^2 + x - 6 \amp = 0\\ (x-2)(x+3) \amp = 0\\ x \amp = 2\\ x \amp = -3 \end{align*}

I have two critical points, \(x=2\) and \(x=-3\text{.}\) The domain of the function is \(\RR\text{,}\) so the intervals are \((-\infty,-3)\text{,}\) \((-3,2)\) and \((2,\infty)\text{.}\) I check the sign of the derivative on these intevals; I do this by simply choosing any point in the interval and evaluated. The following table shows the intervals and the behaviour of the derivative on each interval.

\begin{align*} \amp (-\infty, -3) \amp \amp (-3,2) \amp \amp (2,\infty) \\ \amp f^\prime(-4) = 36 \amp \amp f^\prime(0) = -36 \amp \amp f^\prime(3) = 36 \\ \amp f^\prime(-4) \gt 0 \amp \amp f^\prime(0) \lt 0 \amp \amp f^\prime(3) \gt 0 \\ \amp \text{increasing} \amp \amp \text{decreasing} \amp \amp \text{increasing} \end{align*}

Then at \(x=-3\text{,}\) the function switches from increasing to decreasing, so \((-3,85)\) is a local maximum. At \(x=2\text{,}\) the function switches from decreasing to increasing, so \((2,-48)\) is a local minimum. Figure 10.1.4 shows the behaviour.

Figure 10.1.4. Extrema of a Cubic

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