The Extreme Value Theorem

(a) Yes. The polynomial $f(x) = x^2 - 4x + 3$ is continuous everywhere, and $[0, 5]$ is a closed interval. EVT applies.

(b) No. Although $[0, \pi/2]$ is written as closed, $\tan x$ has a vertical asymptote at $x = \pi/2$. The function is not continuous on this interval (it’s undefined at the right endpoint). EVT does not apply.

(c) Yes. The function $h(x) = \sqrt{x}$ is continuous for $x \geq 0$, so it’s continuous on $[1, 9]$. The interval is closed. EVT applies.

(d) No. The interval $(0, 1]$ is not closed (missing the left endpoint). Even though $1/x$ is continuous on $(0, 1]$, EVT does not apply.

(a) The EVT requires $f$ to be continuous on the closed interval $[0, 3]$. Since $f(x) = 1/(x-1)$ is undefined at $x = 1$, the function is not continuous on $[0, 3]$. The hypothesis fails, so the EVT does not apply.

(b) No absolute maximum exists. As $x \to 1^+$, we have $f(x) \to +\infty$. The function takes arbitrarily large positive values, so there is no largest value.

(c) No absolute minimum exists. As $x \to 1^-$, we have $f(x) \to -\infty$. The function takes arbitrarily large negative values (arbitrarily small), so there is no smallest value.

This example shows that violating continuity can cause both the maximum and minimum to fail to exist.

(a) The function $f(x) = x^3 - 6x^2 + 9x + 2$ is a polynomial, so it is continuous everywhere, in particular on $[0, 4]$. Since $[0, 4]$ is a closed interval and $f$ is continuous on it, the Extreme Value Theorem guarantees that $f$ attains both an absolute maximum and an absolute minimum somewhere in $[0, 4]$.

(b) The absolute extrema can occur only at:

1. Endpoints: $x = 0$ and $x = 4$
2. Critical numbers: values of $x$ in $(0, 4)$ where $f'(x) = 0$ or $f'(x)$ does not exist

Since $f$ is a polynomial, $f'(x)$ exists everywhere, so we only need to find where $f'(x) = 0$.

(c) Endpoint values:

• $f(0) = 0 - 0 + 0 + 2 = 2$
• $f(4) = 64 - 96 + 36 + 2 = 6$

No, we cannot determine the absolute extrema from endpoints alone. The EVT tells us extrema exist, but they might occur at interior critical points. We need to find where $f'(x) = 0$ and compare those values to the endpoint values.

(Note: $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$, so critical numbers are $x = 1$ and $x = 3$. We’d need to evaluate $f(1)$ and $f(3)$ to complete the analysis.)

(a) For any $\epsilon > 0$, the function $f(x) = \ln x$ is continuous on $[\epsilon, e]$, and EVT applies there.

There is no single largest closed interval contained in $(0, e]$ because any closed interval $[a, e]$ with $a > 0$ is contained in $(0, e]$, but we can always find a larger one by choosing smaller $a$.

However, for practical purposes, on any interval $[\epsilon, e]$ with $\epsilon > 0$:

• EVT guarantees an absolute max: $f(e) = 1$ (since $\ln$ is increasing)
• EVT guarantees an absolute min: $f(\epsilon) = \ln \epsilon$

(b) The function $g(x) = x/(x^2-4)$ has discontinuities at $x = -2$ and $x = 2$.

The largest closed intervals within $[-3, 3]$ on which $g$ is continuous are:

• $[-3, -2)$: not closed on right
• $(-2, 2)$: not closed
• $(2, 3]$: not closed on left

So we must exclude the discontinuities. The three largest closed intervals are:

1. $[-3, -2 - \epsilon]$ for small $\epsilon > 0$
2. $[-2 + \epsilon, 2 - \epsilon]$ for small $\epsilon > 0$
3. $[2 + \epsilon, 3]$ for small $\epsilon > 0$

On each such interval, EVT guarantees both an absolute maximum and an absolute minimum exist.

(a) Counterexample:

Define $f(x) = \begin{cases} 1 & \text{if } x = 0 \\ x & \text{if } 0 < x \leq 1 \end{cases}$ on $[0, 1]$.

• $f$ has absolute maximum $f(1) = 1$ at $x = 1$ (and also at $x = 0$)
• $f$ has absolute minimum approaching 0 as $x \to 0^+$...

Wait, that’s not quite right. Try again.

Better counterexample:

Define $f(x) = \begin{cases} 0 & \text{if } x = 0 \\ 1 & \text{if } 0 < x < 1 \\ 2 & \text{if } x = 1 \end{cases}$ on $[0, 1]$.

• Absolute minimum: $f(0) = 0$ at $x = 0$ ✓
• Absolute maximum: $f(1) = 2$ at $x = 1$ ✓
• But $f$ is discontinuous at both $x = 0$ and $x = 1$

This proves the converse is false: a function can have absolute extrema without being continuous.

(b) Proof:

We show that $g$ is continuous on $[a, b]$:

1. On $(a, b)$: $g(x) = f(x)$, and $f$ is continuous on $(a, b)$ by hypothesis. So $g$ is continuous on $(a, b)$.

2. At $x = a$: $$\lim_{x \to a^+} g(x) = \lim_{x \to a^+} f(x) = L = g(a)$$ So $g$ is right-continuous at $a$.

3. At $x = b$: $$\lim_{x \to b^-} g(x) = \lim_{x \to b^-} f(x) = L = g(b)$$ So $g$ is left-continuous at $b$.

Therefore, $g$ is continuous on the closed interval $[a, b]$.

By the Extreme Value Theorem, $g$ attains an absolute maximum and an absolute minimum on $[a, b]$. $\square$

(c) Example:

$f(x) = \cos(2\pi x)$ on $[0, 2]$.

• $f$ is continuous on $[0, 2]$ (cosine is continuous everywhere)
• The absolute maximum value is 1
• This maximum is attained at $x = 0, 1, 2$ (three points on this interval)

For infinitely many points: $f(x) = \cos(2\pi x)$ on $[0, 100]$ attains its maximum of 1 at $x = 0, 1, 2, \ldots, 100$ (101 points).

For truly infinitely many: $f(x) = 1$ (constant function) on $[0, 1]$ attains its maximum of 1 at every point in $[0, 1]$, uncountably many points.