Fermat's Theorem and Absolute Extrema

We shall go on to the end, we shall fight in France, we shall fight on the seas and oceans, we shall fight with growing confidence and growing strength in the air, we shall defend our island, whatever the cost may be. [···] We shall fight on the beaches, we shall fight on the landing grounds, we shall fight in the fields and in the streets, we shall fight in the hills; we shall never surrender, Winston Churchill.

Core Calculus Definitions

Extrema (Maxima and Minima)

The maxima and minima of a function, known collectively as extrema, are the largest and smallest values of the function. They fall into two categories:

• Local (or Relative) Extrema: The largest or smallest value within a specific, localized range.
• Absolute (or Global) Extrema: The single largest or smallest value on the entire domain of the function.

Key Distinction: The definition of absolute extrema entails that an absolute extremum, unlike a relative extremum, may fall on an endpoint of a closed interval.

Critical Points

A critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero ($f'(x) = 0$). The value of the function at a critical point (the $y$-value) is called a critical value.

The squeeze or sandwich theorem

The squeeze or sandwich theorem. If a function f lies between two functions g and h, and the limits of each of them at a particular point are both equal to L, then the limit of f at that particular point is also equal to L.

Formal definition. Let f, g, and h be functions defined on an interval I that contains x = a, except possibly at x = a itself. Suppose that for every x in I not equal to a, we have $g(x) \le f(x) \le h(x)$, and also suppose that $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L.$ Then, it must also be true that: $\lim_{x \to a} f(x) = L.$

The Extreme Value Theorem

The Extreme Value Theorem. If f is a continuous function on an closed interval $[a, b]$, then f has both maximum and minimum values on $[a, b]$. It states that if a function is continuous on a closed bounded interval, then it is guaranteed to hit a highest point and a lowest point somewhere within that interval (either at a peak/valley or at the endpoints a or b).

Fermat’s Theorem

Fermat’s Theorem. If a function $f(x)$ is defined on the interval $(a, b)$, has a local extremum (a relative maximum or minimum) at $x = c$, and $f'(c)$ exists, then $x = c$ is a critical point of $f(x)$ and $f'(c) = 0$. (Figure 1.a.)

Proof:

Let’s assume that $f(x)$ has a relative maximum on the interval at $x = c$. The proof for a relative minimum is nearly identical, so there is no loss of generality here.

By the definition of the derivative: $f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$

Because $c$ is a relative maximum, we know that $f(c) \ge f(c+h)$ for all values of $h$ close to $0$. This means the numerator, $f(c+h) - f(c)$, must be $\le 0$.

Since we assume the limit exists, we can evaluate the one-sided limits:

1. Approaching from the right ($h > 0$): $\lim_{h \to 0^{+}} \frac{f(c+h) - f(c)}{h} \le 0$ because we are dividing a negative or zero number by a positive number, so the result is $\le 0$.
2. Approaching from the left ($h < 0$): $\lim_{h \to 0^{-}} \frac{f(c+h) - f(c)}{h} \ge 0$ because we are dividing a negative or zero number by a negative number, so the result is $\ge 0$.

For the overall two-sided limit to exist, both one-sided limits must equal the same value. The only number that is both $\le 0$ and $\ge 0$ is $0$. Therefore, $f'(c) = 0$. ∎

Important Nuances and Extreme Cases

The Converse is Not True

Just because $f'(c) = 0$ does not automatically mean that $c$ is a local maximum or minimum. We can have a critical point c such that f(c) is not a local maximum or minimum.

Consider the function $f(x) = x^3$ (Figure i). The derivative is $f'(x) = 3x^2$, which means $f'(0) = 0$. However, the point $(0, 0)$ is neither a local maximum nor a minimum; it is simply an inflection point where the graph briefly flattens out before continuing upward.

Extreme Cases: Constant Functions

For a constant function like $f(x) = c$ (Figure ii):

• Every point is critical because $f'(x) = 0$ for all $x \in \mathbb{R}$.
• Every point is simultaneously a global and local maximum and minimum, and y = c is both the minimum and maximum value.

Piecewise Function Example

Consider this piecewise function (Figure iii):

$f(x) = \begin{cases} \vert{}x\vert{}, & x < 1 \\ -(x - 2)^2 + 2, & x \ge 1 \end{cases}$

Its derivative is:

$f'(x) = \begin{cases} -1, & x < 0 \\ 1, & 0 < x < 1 \\ -2(x - 2), & x > 1 \end{cases}$

This function has three critical points:

• $x = 0$: $f'(0)$ is undefined (a sharp corner in the absolute value graph). This is a local minimum.
• $x = 1$: $f'(1)$ is undefined (the two pieces connect, but their slopes don’t match).
• $x = 2$: $f'(2) = 0$ because $-2(x-2) = 0 \implies x = 2$. This is a local maximum.

  2 is a local maximum. Since f’(2) exists, we already knew that f’(2) = 0.

How to Find Absolute Extrema on a Closed Interval $[a, b]$

If a function $f$ is continuous on a closed interval $[a, b]$, the Extreme Value Theorem guarantees that $f$ has both an absolute maximum and an absolute minimum value on that interval.

By Fermat’s theorem, every local extremum in the interior of a differentiable function is necessarily a critical point. Therefore, to find the absolute extremes, we only need to check the critical points and the boundaries.

1. Verify Continuity. Confirm that the function is continuous on the entire closed interval $[a, b]$.
2. Find Critical Values. Calculate the derivative $f'(x)$ and determine all critical points (where $f'(x) = 0$ or is undefined) that fall inside the interval $(a, b)$.
3. Evaluate the Function. Plug the $x$-values of your critical points, along with the endpoints $a$ and $b$, back into the original function $f(x)$.
4. Identify the Extrema. Compare the resulting $y$-values. The largest $y$-value is your absolute maximum, and the smallest $y$-value is your absolute minimum.

Solved Exercises

Extrema with no critical points in the interval

Find the maximum value of $f(x) = x^2 - 7x + 12$ on the interval $[0, 1]$ (Figure iv).

1. $f(x)$ is a polynomial, so it is continuous and differentiable everywhere.
2. Find critical points: $f'(x) = 2x - 7 = 0 \implies 2x = 7 \implies x = 3.5$.
3. Since $3.5$ is not in our interval $[0, 1]$, there are no critical points to check.
4. We only evaluate the endpoints: $f(0) = 12$, $f(1) = 1^2 - 7(1) + 12 = 6$
5. Result: The absolute maximum value on this interval is $12$ (occurring at $x = 0$).

Extrema where the derivative is undefined

Find any absolute extrema for $f(x) = \vert{}x\vert{}$ on the interval $[-1, 1]$ (Figure 1).

1. $f(x)$ is continuous, so the EVT applies.
2. The only critical point is at $x = 0$, because $f'(0)$ is undefined (it’s a sharp corner).
3. Evaluate the critical point and the endpoints: $f(-1) = \vert{}-1\vert{} = 1$, $f(0) = \vert{}0\vert{} = 0$, and $f(1) = \vert{}1\vert{} = 1$
4. Result: The absolute maximums are at $x = \pm 1$ ($y=1$), and the absolute minimum is at $x = 0$ ($y=0$). This shows that an absolute extremum can occur where a derivative does not exist.

In Figures 1-5, we see several different possibilities for critical points. In some cases, the functions have local extrema (maximum -1-, -2- or minimum -3-) at critical points, whereas in other cases (4, 5) the functions do not.

Extrema with a non-zero derivative

Find any absolute extrema for $f(x) = x$ on the interval $[1, 2]$ (Figure 2).

1. $f(x)$ is continuous.
2. $f'(x) = 1$. Since $1 \neq 0$, there are no critical points.
3. Evaluate endpoints: $f(1) = 1$ and $f(2) = 2$
4. Result: The absolute minimum is $1$ (at $x=1$) and the absolute maximum is $2$ (at $x=2$).

Extrema with multiple critical points

Find any absolute extrema for $f(x) = 3x^4 - 4x^3$ on the interval $[-1, 2]$ (Figure 3).

1. $f(x)$ is continuous.
2. Find critical points: $f'(x) = 12x^3 - 12x^2 = 12x^2(x - 1) = 0$. This gives us critical points at $x = 0$ and $x = 1$. Both are within our interval.
3. Evaluate critical points and endpoints: $f(-1) = 3(-1)^4 - 4(-1)^3 = 3 + 4 = 7$, $f(0) = 3(0)^4 - 4(0)^3 = 0$, $f(1) = 3(1)^4 - 4(1)^3 = 3 - 4 = -1$, and $f(2) = 3(2)^4 - 4(2)^3 = 48 - 32 = 16$
4. Result: $f(x)$ has an absolute maximum of $16$ (at $x=2$) and an absolute minimum of $-1$ (at $x=1$). Notice that $x=0$ is a critical point, but it is neither the absolute maximum nor minimum —just a brief flattening out of the curve.

When the Extreme Value Theorem Fails

Find any absolute extrema for $f(x) = \frac{1}{x}$ on the interval $[-1, 1]$ (Figure 4).

The function $f(x) = \frac{1}{x}$ is not continuous at $x = 0$, which falls inside our closed interval $[-1, 1]$. Because continuity is broken, we cannot apply the Extreme Value Theorem.

Furthermore, investigating the behavior around $x = 0$ reveals vertical asymptotes: $\lim_{x \to 0^+} \frac{1}{x} = \infty$, $\lim_{x \to 0^-} \frac{1}{x} = -\infty$.

Because the function grows infinitely large in both the positive and negative directions within the interval, $f(x)$ does not have any absolute extrema on $[-1, 1]$.

A Standard Polynomial Example

Problem: Find any absolute extrema for $f(x) = \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x$ on the interval $[0, 5]$ (Figure 5).

1. Because $f(x)$ is a polynomial, it is continuous and differentiable everywhere. We can safely apply the Extreme Value Theorem.
2. Find critical points: Set the derivative to zero. $f'(x) = x^2 - 5x + 4 = 0$. $(x - 4)(x - 1) = 0$
   The critical points are $x = 1$ and $x = 4$. Both are inside our interval $[0, 5]$.
3. Evaluate at critical values and endpoints: $f(0) = 0$, $f(1) = \frac{1}{3}(1)^3 - \frac{5}{2}(1)^2 + 4(1) = \frac{11}{6} \approx 1.83$, $f(4) = \frac{1}{3}(4)^3 - \frac{5}{2}(4)^2 + 4(4) = -\frac{8}{3} \approx -2.66$, and $f(5) = \frac{1}{3}(5)^3 - \frac{5}{2}(5)^2 + 4(5) = -\frac{5}{6} \approx -0.83$
4. Comparing the $y$-values, the function has an absolute maximum at $x=1$ ($y \approx 1.83$) and an absolute minimum at $x=4$ ($y \approx -2.66$).

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