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Module III — Extreme Values and Mean Value Theorem

Module III — Extreme Values and the Mean Value Theorem

Course: MAT1CJ101 — Differential Calculus
Hours: 11 | Textbook Sections: §3.1, §3.2, §3.3 (Thomas & Finney, 9th Ed.)

1. Extreme Values of Functions

1.1 Absolute and Local Extrema

Definitions:

Let $f$ be defined on a domain $D$.

  • Absolute maximum: $f(c)$ is the absolute maximum of $f$ on $D$ if $f(c) \geq f(x)$ for all $x$ in $D$.
  • Absolute minimum: $f(c)$ is the absolute minimum of $f$ on $D$ if $f(c) \leq f(x)$ for all $x$ in $D$.
  • Local maximum: $f(c) \geq f(x)$ for all $x$ near $c$ (in some open interval around $c$).
  • Local minimum: $f(c) \leq f(x)$ for all $x$ near $c$.

Extrema = maxima and minima (both absolute and local).

Note: An absolute extremum is also a local extremum (if it occurs in the interior of the domain). But a local extremum need not be absolute.

1.2 The Extreme Value Theorem

Theorem (EVT): If $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a, b]$.

Why the conditions matter:

  • Without continuity: $f$ might jump over the maximum value.
  • Without a closed interval: $f(x) = x$ on $(0,1)$ has no maximum (approaches but never reaches 1).
  • Without a bounded interval: $f(x) = x$ on $[0, \infty)$ has no maximum.

1.3 Where Can Extrema Occur?

Theorem: If $f$ has a local extremum at an interior point $c$ and $f’(c)$ exists, then $f’(c) = 0$.

Proof sketch: If $f$ has a local max at $c$, then $f(x) \leq f(c)$ near $c$. For $h > 0$ small, $\dfrac{f(c+h)-f(c)}{h} \leq 0$, so the right-hand limit $\leq 0$. For $h < 0$ small, $\dfrac{f(c+h)-f(c)}{h} \geq 0$, so the left-hand limit $\geq 0$. Both limits equal $f’(c)$, so $f’(c) = 0$.

Critical points of $f$ are interior points $c$ where either:

  • $f’(c) = 0$ (stationary point), or
  • $f’(c)$ does not exist.

Key fact: Every local extremum at an interior point is a critical point. But not every critical point is a local extremum.

Worked Example 1: Find the critical points of $f(x) = x^3 - 3x + 1$.

\[f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)\] \[f'(x) = 0 \implies x = 1 \text{ or } x = -1\]

Critical points: $x = 1$ and $x = -1$. \(f(1) = 1 - 3 + 1 = -1, \qquad f(-1) = -1 + 3 + 1 = 3\)

2. Finding Extrema — The Closed Interval Method

To find the absolute extrema of a continuous function $f$ on $[a, b]$:

  1. Find all critical points of $f$ in $(a, b)$.
  2. Evaluate $f$ at each critical point.
  3. Evaluate $f$ at the endpoints: $f(a)$ and $f(b)$.
  4. The largest value among all those computed is the absolute maximum; the smallest is the absolute minimum.

Worked Example 2: Find the absolute extrema of $f(x) = x^2 - 3$ on $[-2, 3)$. (Oct 2024, Q7)

Note: the interval is half-open $[-2, 3)$ — the right endpoint is not included.

\[f'(x) = 2x = 0 \implies x = 0\]

Critical point: $x = 0$, $f(0) = -3$. Endpoint: $x = -2$, $f(-2) = 4 - 3 = 1$. Right end $x = 3$ is not included; $f(x) \to 6$ as $x \to 3^-$ but 6 is never attained.

Absolute minimum: $f(0) = -3$ (at $x = 0$). Absolute maximum: does not exist! The function approaches 6 but never reaches it.

(This illustrates why EVT requires a closed interval.)

Worked Example 3: Find the absolute extrema of $f(x) = 2x^3 - 3x^2 - 12x + 1$ on $[-2, 3]$.

\[f'(x) = 6x^2 - 6x - 12 = 6(x^2-x-2) = 6(x-2)(x+1)\]

$f’(x) = 0 \implies x = 2$ or $x = -1$ (both in $[-2, 3]$) ✓

Evaluate at critical points and endpoints:

\(f(-2) = 2(-8) - 3(4) - 12(-2) + 1 = -16 - 12 + 24 + 1 = -3\) \(f(-1) = 2(-1) - 3(1) - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8\) \(f(2) = 2(8) - 3(4) - 12(2) + 1 = 16 - 12 - 24 + 1 = -19\) \(f(3) = 2(27) - 3(9) - 12(3) + 1 = 54 - 27 - 36 + 1 = -8\)

Absolute maximum: $f(-1) = 8$ Absolute minimum: $f(2) = -19$

3. The Mean Value Theorem

3.1 Rolle’s Theorem

Theorem (Rolle): Let $f$ be continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$. Then there exists at least one $c \in (a, b)$ with $f’(c) = 0$.

Intuition: If you start and end at the same height, at some point in between you must be going neither up nor down (instantaneous velocity $= 0$).

Worked Example 4: Verify Rolle’s theorem for $f(x) = x^2 - 2x$ on $[0, 2]$.

$f$ is a polynomial (continuous everywhere, differentiable everywhere). \(f(0) = 0, \qquad f(2) = 4 - 4 = 0 \quad \text{✓ Equal endpoints.}\)

\[f'(x) = 2x - 2 = 0 \implies x = 1, \text{ which lies in } (0, 2). ✓\]

Rolle guarantees $c = 1$.

3.2 The Mean Value Theorem (MVT)

Theorem (MVT): Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists at least one $c \in (a, b)$ such that:

\[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Intuition: The instantaneous rate of change equals the average rate of change at least once. The tangent line at some interior point is parallel to the secant line from $a$ to $b$.

How MVT follows from Rolle: Define $g(x) = f(x) - L(x)$, where $L(x)$ is the secant line from $(a, f(a))$ to $(b, f(b))$. Then $g(a) = g(b) = 0$ and $g$ is differentiable. Rolle gives $g’(c) = 0$, which means $f’(c) = L’(c) = \dfrac{f(b)-f(a)}{b-a}$.

Worked Example 5: Find $c$ guaranteed by MVT for $f(x) = x^2$ on $[1, 3]$.

\[\text{Average rate} = \frac{f(3) - f(1)}{3 - 1} = \frac{9-1}{2} = 4\]

MVT says $f’(c) = 4$. Since $f’(x) = 2x$: \(2c = 4 \implies c = 2\)

$c = 2$ lies in $(1, 3)$. ✓

3.3 Consequences of the MVT

Corollary 1 (Zero Derivative → Constant): If $f’(x) = 0$ for all $x$ in an interval $I$, then $f$ is constant on $I$. (Oct 2024, Q8)

Proof: Take any $a, b$ in $I$ with $a < b$. By MVT, $f(b) - f(a) = f’(c)(b-a) = 0 \cdot (b-a) = 0$. So $f(b) = f(a)$. Since $a, b$ were arbitrary, $f$ is constant.

Corollary 2 (Equal Derivatives → Differ by Constant): If $f’(x) = g’(x)$ on $I$, then $f(x) = g(x) + C$ for some constant $C$.

Corollary 3 (Sign of Derivative → Monotonicity):

  • $f’(x) > 0$ on $(a, b)$ → $f$ is increasing on $(a, b)$
  • $f’(x) < 0$ on $(a, b)$ → $f$ is decreasing on $(a, b)$

Proof: For $a < x_1 < x_2 < b$, MVT gives $f(x_2) - f(x_1) = f’(c)(x_2 - x_1)$ for some $c$. If $f’ > 0$, the right side is positive, so $f(x_2) > f(x_1)$.

3.4 Functions with the Same Derivative — Antiderivatives

If $F’(x) = f(x)$, then $F$ is called an antiderivative of $f$. Corollary 2 says all antiderivatives of $f$ differ by a constant:

\[\text{All antiderivatives of } f(x) = 2x \text{ are } F(x) = x^2 + C, \quad C \in \mathbb{R}\]

4. Increasing and Decreasing Functions

4.1 Identifying Intervals of Increase and Decrease

Procedure:

  1. Find $f’(x)$.
  2. Solve $f’(x) = 0$ and find where $f’(x)$ is undefined.
  3. These points divide the domain into intervals.
  4. Test the sign of $f’$ on each interval (plug in a test point).
  5. $f$ is increasing where $f’ > 0$, decreasing where $f’ < 0$.

Worked Example 6: Find where $f(x) = x^3 + 12x + 5$ is increasing/decreasing on $[-3, 3]$. (Oct 2024, Q19a)

\[f'(x) = 3x^2 + 12 = 3(x^2 + 4)\]

Since $x^2 + 4 \geq 4 > 0$ always, $f’(x) > 0$ everywhere on $[-3, 3]$.

$f$ is strictly increasing on $[-3, 3]$.

Since $f$ is strictly monotone, it has no interior local extrema.

Absolute minimum: $f(-3) = -27 - 36 + 5 = -58$ Absolute maximum: $f(3) = 27 + 36 + 5 = 68$

Worked Example 7: Find increasing/decreasing intervals for $f(x) = x^4 - 8x^2$.

\[f'(x) = 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)\]

Critical points: $x = -2, 0, 2$

Sign analysis:

Interval Sign of $f’$ Behavior
$x < -2$ $4(-)(-)(-) = -$ decreasing
$-2 < x < 0$ $4(-)(-)(+) = +$ increasing
$0 < x < 2$ $4(+)(-)(+) = -$ decreasing
$x > 2$ $4(+)(+)(+) = +$ increasing

5. The First Derivative Test

Theorem (First Derivative Test): Suppose $c$ is a critical point of $f$.

Sign of $f’$ Left of $c$ Right of $c$ Conclusion
Changes $-$ to $+$ $f’ < 0$ $f’ > 0$ Local minimum at $c$
Changes $+$ to $-$ $f’ > 0$ $f’ < 0$ Local maximum at $c$
No sign change same sign same sign Not an extremum

Worked Example 8: Classify the critical points of $f(x) = x^4 - 8x^2$.

From Worked Example 7:

  • $x = -2$: $f’$ changes from $-$ to $+$ → local minimum, $f(-2) = 16 - 32 = -16$
  • $x = 0$: $f’$ changes from $+$ to $-$ → local maximum, $f(0) = 0$
  • $x = 2$: $f’$ changes from $-$ to $+$ → local minimum, $f(2) = 16 - 32 = -16$

Worked Example 9: Show $f(x) = x^4 + 3x + 1$ has exactly one zero in $[-2, -1]$. (Oct 2024, Q16)

Existence (IVT): \(f(-2) = 16 - 6 + 1 = 11 > 0\) \(f(-1) = 1 - 3 + 1 = -1 < 0\) $f$ is a polynomial (continuous). By IVT, there is at least one zero in $(-2, -1)$.

Uniqueness (monotonicity): \(f'(x) = 4x^3 + 3\) On $[-2, -1]$: $x \in [-2, -1]$, so $x^3 \in [-8, -1]$. \(4x^3 \in [-32, -4], \text{ so } f'(x) = 4x^3 + 3 \in [-29, -1] < 0\) $f$ is strictly decreasing on $[-2, -1]$.

A strictly decreasing function can cross zero at most once. Therefore, $f$ has exactly one zero in $[-2, -1]$.

Worked Example 10: Find intervals of increase/decrease and local extrema for $f(x) = x^3 + 12x + 5$ on $[-3, 3]$. (Oct 2024, Q19a)

\[f'(x) = 3x^2 + 12 > 0 \text{ everywhere (see Worked Example 6)}\]

$f$ is increasing on $[-3, 3]$ with no local extrema in the interior.

Absolute minimum at $x = -3$: $f(-3) = -58$ Absolute maximum at $x = 3$: $f(3) = 68$

6. No Local Extremum at a Boundary-Like Point

Worked Example 11: Give an example of a function on $[0, 1]$ with no local maximum or minimum at $x = 0$. (Oct 2024, Q9)

Consider $f(x) = x$ on $[0, 1]$.

\[f'(x) = 1 > 0 \text{ everywhere, so } f \text{ is strictly increasing}\]

At $x = 0$: $f(0) = 0$ is the absolute minimum, but by the definition of local min in some textbooks, a boundary point may or may not count.

If “local extremum” requires an open neighbourhood: then $x = 0$ (a boundary point of $[0,1]$) does not have a local min in the interior sense.

Alternatively: $f(x) = x$ on $[0,1]$ has $f(0) = 0$ as an absolute minimum but not a “local” minimum because there is no open interval around 0 contained in $[0,1]$.

7. Summary

Concept Key Statement
EVT Continuous $f$ on $[a,b]$ attains its absolute max and min
Critical point Interior $c$ where $f’(c) = 0$ or $f’(c)$ DNE
Closed interval method Evaluate $f$ at critical points + endpoints
Rolle’s Theorem $f(a) = f(b) \to \exists c: f’(c) = 0$
MVT $\exists c: f’(c) = \dfrac{f(b)-f(a)}{b-a}$
$f’ = 0$ everywhere $f$ is constant (Corollary 1)
$f’ > 0$ on interval $f$ increasing; $f’ < 0 \to$ decreasing
First Derivative Test Sign change of $f’$ at critical point → local extremum

8. Practice Problems

Sec A style (3 marks each):

  1. Find all critical points of $f(x) = x^3 - 6x^2 + 9x - 2$.

  2. Find the absolute extrema of $f(x) = x^2 - 3$ on $[-2, 3)$. (Oct 2024, Q7)

  3. If $f’(x) = 0$ for all $x$ in an interval $I$, prove $f$ is constant on $I$. (Oct 2024, Q8)

  4. Give an example of a function on $[0, 1]$ with no local maximum or minimum value at $0$. Explain. (Oct 2024, Q9)

  5. Verify Rolle’s Theorem for $f(x) = x(x-1)(x-2)$ on $[0, 2]$.

  6. Find $c$ given by MVT for $f(x) = x^3$ on $[0, 2]$.

Sec B style (6 marks each):

  1. Find the intervals on which $f(x) = x^3 + 12x + 5$, $-3 \leq x \leq 3$ is increasing and decreasing. Find all extreme values. (Oct 2024, Q19a)

  2. Show $f(x) = x^4 + 3x + 1$ has exactly one zero in $[-2, -1]$. (Oct 2024, Q16)

  3. Find all local maxima and minima of $f(x) = x^4 - 8x^2 + 5$. Use the First Derivative Test.

  4. A function $f$ satisfies $f(1) = 3$ and $f’(x) \leq 2$ for all $x \in [1, 4]$. What is the largest $f(4)$ can be? Justify using MVT.

Sec C style (10 marks):

  1. Let $f(x) = x^3 - 3x^2 - 9x + 5$ on $[-2, 6]$. (a) Find all critical points. Classify them using the First Derivative Test. (b) Find the absolute maximum and minimum on $[-2, 6]$. (c) Sketch a rough graph showing the increasing/decreasing intervals and extrema.