Local extrema

Quite often, expressions such as "minimize costs" or "maximize space," "minimize" and "maximize" are usually interpreted as synonyms of "decrease" and "increase," respectively. However, it goes beyond that, as they refer to particular values of given functions. In this topic, we will define these extreme values and learn how to use derivatives to properly identify them.

Extrema

In general terms, a minimum is the smallest value a function $f$ can output, while maximum refers to the largest value. Together, they are commonly referred to as extrema.

Given our function $f$ and an arbitrarily small positive value $\delta > 0$, we can define local extrema as follows:

• $f$ has a local minimum at $x_0$ if $f(x_0) \leq f(x)$ for all $x \in (x_0-\delta, x_0+\delta)$
• $f$ has a local maximum at $x_0$ if $f(x_0) \geq f(x)$ for all $x \in (x_0-\delta, x_0+\delta)$

We can estimate the local extrema of some functions simply by inspecting their graphs. However, in order to precisely determine these points, we need to use derivatives.

Critical points

If $f(x_0)$ is a local minimum for $x \in (x_0-\delta, x_0+\delta)$, then $f(x)$ decreases on the interval $(x_0-\delta,x_0)$ and increases on $(x_0,x_0+\delta)$. So $x = x_0$ is the point where the function switches from decreasing to increasing, or vice versa if $f(x_0)$ is a local maximum instead. We can use this insight to find local extrema analytically.

We already know how to recognize when a function $f$ is either decreasing or increasing at a given point. We also know that the derivative $f'(x_0)$ represents the slope of the line tangent to $f(x)$ at the point $x_0$. When $f'$ is positive, the tangent line points up; when it's negative, the tangent line points down. So,

• if $f'(x_0) > 0$, then $f(x)$ is increasing at $x = x_0$

• if $f'(x_0) < 0$, then $f(x)$ is decreasing at $x = x_0$

• if $f'(x_0) = 0$, then $x = x_0$ is a critical point of $f(x)$. The graph of the function is momentarily flat. This covers several possibilities: $x_0$ could be a local minimum, a local maximum, or something else entirely. We'll need more information to determine exactly what's happening.

For example, let's say

$$f(x) = \frac{1}{3}x^3 - x \implies f'(x)=x^2 -1$$At $x = 2$, we have $f'(2)=(2)^2 -1=3$.

At $x = 1/2$, we have $f'(\frac{1}{2})=(\frac{1}{2})^2 -1=-\frac{3}{4}$.

At $x = 1$, we have $f'(1)=(1)^2 -1=0$.

Therefore, we can say that $f(x)$ is increasing at $x = 2$ and decreasing at $x = 1/2$, and it has a critical point at $x=1$.

So how do we find out if our critical point is a local extremum?

Second-derivative test

Just like the first derivative lets us know whether $f(x)$ is increasing or decreasing, the second derivative tells us whether $f'(x)$ is increasing or decreasing. If the second derivative is positive at a critical point $x = x_0$, then the first derivative (slope of the tangent line) is going from negative to positive, and $f(x_0)$ is therefore a local minimum of the function. Likewise a negative second derivative at the critical point implies that we have a local maximum.

Then, we have the following:

• if $f'(x_0) = 0$ and $f''(x_0) > 0$, then $f(x)$ has a local minimum at $x = x_0$;

• if $f'(x_0) = 0$ and $f''(x_0) < 0$, then $f(x)$ has a local maximum at $x = x_0$;

• if $f'(x_0) = 0$ and $f''(x_0) = 0$, then the second derivative test has failed to tell us anything. There could be a local minimum, local maximum, or neither at $x = x_0$.

To continue with our example, we have

$$f(x) = \frac{1}{3}x^3 - x \implies f'(x)=x^2 -1 \implies f''(x)=2x$$At $x = 1$, we have

$$f''(1)=2\cdot(1)=2 > 0$$

Therefore, our critical point at $x = 1$ is a local minimum.

We can use the first-derivative test to find all the critical points of a function, then the second-derivative test to find out which of them refer to local minima or local maxima.

Taking our example function $f(x) = \frac{1}{3}x^3 - x$, we have:

$$f'(x)=x^2 -1 =0 \implies x=\pm1$$Thus our critical points are $\left(-1, \frac{2}{3}\right)$ and $\left(1, -\frac{2}{3}\right)$.

We also have:

$$f''(1)=2\cdot(1)=2 > 0 \quad ; \quad f''(-1)=2\cdot(-1)=-2 < 0$$So, we have a minimum at $\left(1, -\frac{2}{3}\right)$ and a maximum at $\left(-1, \frac{2}{3}\right)$.

In the previous graph, we can see that if we were to look only in the interval $[-0.5,1.5]$, $f(x)$ wouldn't have a local maximum. Thus, given an interval $[a,b]$, discard those critical points where $x \notin [a,b]$ before applying the second-derivative test.

Global extrema and local extrema

Let's say

$$f(x)=x^4 - x^2$$Then, if we set $f'(x)=0$,

$$f'(x)=4x^3-2x=0 \implies x= \begin{cases} 0\\ \pm\frac{1}{\sqrt{2}} \end{cases}$$We can see that $f(x)$ has three critical points: $\left (-\frac{1}{\sqrt{2}}, -\frac{1}{4}\right)$, $(0, 0)$, and $\left (\frac{1}{\sqrt{2}}, -\frac{1}{4}\right)$.

Then, we have

$$f'(x)=4x^3-2x=0 \implies f''(x)=12x^2-2 \implies \begin{cases} f'' \left (-\frac{1}{\sqrt{2}} \right ) &> 0 \\f''(0)&<0\\ f'' \left (\frac{1}{\sqrt{2}} \right ) &> 0\end{cases}$$Thus, we have two local minima at $\left (\pm \frac{1}{\sqrt{2}}, - \frac{1}{4}\right )$ and a local maximum at $(0, 0)$

We can see that even though $f(0) = 0$ is a local maximum, it is not the largest value in the interval pictured. These values are known as absolute or global extrema. In order to find the global maximum and global minimum of a function in a given interval $[a,b]$, the function must be evaluated at the interval boundaries in order to compare $f(a)$ and $f(b)$ to the local extrema obtained from the second derivative test.

For example, take the same $f(x)=x^4 - x^2$ and the interval $[0,1.5]$. Since $-\frac{1}{\sqrt{2}} <0$, we are left with only two critical points in the interval: $(0, 0)$ and $\left (\frac{1}{\sqrt{2}}, -\frac{1}{4}\right)$.

Now, we evaluate $f(x)$ at the boundaries (endpoints) of the interval:

$$\begin{cases} x = 0 \implies f(0)=0\\ x = 1.5 \implies f(1.5)=2.81 \end{cases}$$Since $f(1.5) > f(0)>f\left(\frac{1}{\sqrt{2}} \right)$, we conclude that $f(x)$ has a global maximum at $x = 1.5$, and a global minimum at $x = \frac{1}{\sqrt{2}} \approx0.707$. We only need to compare three points because $x = 0$ is both a critical point of the function and an endpoint of the interval.

Conclusion

To sum up:

• The least value $f(x_0)$ a function $f$ can output for all $x$ in a small interval near $x_0$ is called a local minimum, while the largest value is called a local maximum. These points are known together as local extrema.
• We can use the first-derivative test to know if $f$ is increasing or decreasing at a given point. A critical point is one where the function is defined, but it is neither increasing nor decreasing. By setting the first derivative equal to zero, we can determine the critical points of $f$.
• We can use the second-derivative test to determine if a critical point is a local extremum.
• In order to define the absolute extrema for a given interval, we have to evaluate $f$ at the boundaries of the interval and compare these values to the local extrema found using the second-derivative test.