# Unit 5: Analytical Applications of Differentiation

# 1. INTRODUCTION

Stop just crunching numbers and start reading the vibe of the graph. Unit 5 is like an autopsy for functions—we're using derivatives to uncover a curve’s hidden "DNA." Whether you’re trying to find the peak G-force on a coaster or maximizing profit without the burnout, this is where calculus gets real. We’re moving from "how to derive" to "where it actually matters."

# 2. ALL KEY CONCEPTS, TERMS, AND PRINCIPLES

# Foundational Theorems

• Extreme Value Theorem (EVT): If a function $f$ is continuous on a closed interval $[a, b]$, then $f$ must attain an absolute maximum and an absolute minimum on that interval.
• Mean Value Theorem (MVT): If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists at least one point $c$ in $(a, b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
• Rolle's Theorem: A specific case of MVT where $f(a) = f(b)$, implying there exists a $c$ such that $f'(c) = 0$.

# Extrema and Critical Points

• Critical Point (or Critical Number): A value $c$ in the domain of $f$ such that $f'(c) = 0$ or $f'(c)$ is undefined.
• Relative (Local) Extrema: Points where a function reaches a peak (relative maximum) or a valley (relative minimum) within a specific neighborhood.
• Absolute (Global) Extrema: The highest or lowest value of a function over its entire domain or a specified interval.

# Function Behavior

• Monotonicity: The characteristic of a function being entirely increasing or entirely decreasing.
  • Increasing: $f'(x) > 0$
  • Decreasing: $f'(x) < 0$
• Concavity: The "bend" of a curve.
  • Concave Up: The rate of change ($f'$) is increasing; $f''(x) > 0$. Visually like a cup $\cup$.
  • Concave Down: The rate of change ($f'$) is decreasing; $f''(x) < 0$. Visually like a frown $\cap$.
• Point of Inflection: A point on a curve where the concavity changes (from up to down or vice versa). This occurs where $f''(x) = 0$ or is undefined, provided the sign of $f''$ actually switches.

# Optimization

• Objective Function: The equation representing the quantity you want to maximize or minimize (e.g., Area, Volume, Cost).
• Constraint: A secondary equation that limits the variables in the objective function (e.g., a fixed amount of fencing or a specific volume).

# 3. IN-DEPTH EXPLANATION

# 3.1 The Mean Value Theorem (MVT)

The MVT is one of the most important theoretical tools in calculus. It bridges the gap between the "average rate of change" and the "instantaneous rate of change."

The Definition:
If $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a $c \in (a, b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

{
  "type": "line",
  "title": "Mean Value Theorem Illustration",
  "subtitle": "f(x) = x² - 4x + 3 on [1, 4]",
  "xLabel": "x",
  "yLabel": "f(x)",
  "labels": ["1.0", "1.5", "2.0", "2.5", "3.0", "3.5", "4.0"],
  "datasets": [
    { 
      "label": "f(x) = x² - 4x + 3", 
      "data": [0, -0.75, -1, -0.75, 0, 1.25, 3], 
      "tension": 0.4 
    },
    { 
      "label": "Secant Line (Avg Rate)", 
      "data": [0, 0.5, 1, 1.5, 2, 2.5, 3], 
      "dashed": true 
    }
  ]
}

Interpretation:
Geometrically, this means there is at least one point on the curve where the tangent line is parallel to the secant line passing through the endpoints $(a, f(a))$ and $(b, f(b))$. In physics, if your average speed on a trip was 60 mph, the MVT guarantees that at least at one moment, your speedometer read exactly 60 mph.

# 3.2 Global Extrema and the Candidates Test

To find the absolute maximum or minimum of a continuous function on a closed interval $[a, b]$, we use the Candidates Test.

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

# 3.3 The First Derivative Test

This test allows us to classify critical points as relative maxima, relative minima, or neither.

• If $f'(x)$ changes from positive to negative at $x = c$, then $f(c)$ is a relative maximum.
• If $f'(x)$ changes from negative to positive at $x = c$, then $f(c)$ is a relative minimum.
• If $f'(x)$ does not change sign, there is no extremum at $x = c$.

# 3.4 Concavity and the Second Derivative Test

Concavity describes the "curvature" of the graph.

• Concave Up: $f'(x)$ is increasing $\implies f''(x) > 0$. The tangent lines lie below the curve.
• Concave Down: $f'(x)$ is decreasing $\implies f''(x) < 0$. The tangent lines lie above the curve.

{
  "type": "line",
  "title": "Visualizing Concavity",
  "subtitle": "f''(x) > 0 (Up) vs f''(x) < 0 (Down)",
  "xLabel": "x",
  "yLabel": "f(x)",
  "labels": ["-2", "-1", "0", "1", "2"],
  "datasets": [
    { 
      "label": "Concave Up (x²)", 
      "data": [4, 1, 0, 1, 4], 
      "tension": 0.4 
    },
    { 
      "label": "Concave Down (-x²)", 
      "data": [-4, -1, 0, -1, -4], 
      "tension": 0.4 
    }
  ]
}

The Second Derivative Test for Relative Extrema:
If $f'(c) = 0$:

1. If $f''(c) < 0$, the graph is concave down, making $f(c)$ a relative maximum.
2. If $f''(c) > 0$, the graph is concave up, making $f(c)$ a relative minimum.
3. If $f''(c) = 0$, the test is inconclusive (use the First Derivative Test).

# 3.5 Optimization Problems

Optimization is the process of finding the "best" value. The workflow for AP Calculus optimization is:

1. Identify variables: Draw a picture if possible.
2. Write the Objective Function: What are you maximizing/minimizing? (e.g., $V = lwh$).
3. Write the Constraint: Use the given info to relate variables (e.g., $2l + 2w = P$).
4. Substitute: Reduce the objective function to a single variable.
5. Differentiate and solve: Find the critical points and justify using the First or Second Derivative Test (or Candidates Test if the domain is closed).

# 4. EXAMPLES

# Example 1: Applying the Mean Value Theorem

Given $f(x) = x^2 - 4x + 3$ on $[1, 4]$, find the value of $c$ that satisfies the MVT.
Solution:

1. $f(x)$ is a polynomial, so it is continuous and differentiable everywhere.
2. Endpoints: $f(1) = 1-4+3 = 0$; $f(4) = 16-16+3 = 3$.
3. Average rate of change: $\frac{3 - 0}{4 - 1} = \frac{3}{3} = 1$.
4. Derivative: $f'(x) = 2x - 4$.
5. Set $f'(c) = 1$:
   $$2c - 4 = 1 \implies 2c = 5 \implies c = 2.5$$
   Since $2.5 \in (1, 4)$, the MVT is satisfied.

# Example 2: Finding Critical Points

Find the critical points of $f(x) = x^3 - 6x^2 + 9x$.
Solution:

1. $f'(x) = 3x^2 - 12x + 9$.
2. Set $f'(x) = 0$: $3(x^2 - 4x + 3) = 0 \implies 3(x-3)(x-1) = 0$.
3. Critical points at $x = 1$ and $x = 3$.

# Example 3: The Candidates Test (Absolute Extrema)

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

1. $f'(x) = 3x^2 - 6x = 3x(x - 2)$.
2. Critical points in interval: $x = 0, x = 2$.
3. Evaluate $f$ at critical points and endpoints:
   • $f(-1) = (-1)^3 - 3(-1)^2 + 1 = -3$
   • $f(0) = 0 - 0 + 1 = 1$
   • $f(2) = 8 - 12 + 1 = -3$
   • $f(3) = 27 - 27 + 1 = 1$
     Result: Absolute Max is $1$ (at $x=0, 3$). Absolute Min is $-3$ (at $x=-1, 2$).

# Example 4: First Derivative Test

Classify the relative extrema of $f(x) = \frac{1}{3}x^3 - x^2 - 3x + 5$.
Solution:

1. $f'(x) = x^2 - 2x - 3 = (x-3)(x+1)$.
2. Critical points: $x = -1, 3$.
3. Sign Chart for $f'$:
   • $(-\infty, -1)$: $f'(-2) = (-)(- ) = +$ (Increasing)
   • $(-1, 3)$: $f'(0) = (-)(+) = -$ (Decreasing)
   • $(3, \infty)$: $f'(4) = (+)(+) = +$ (Increasing)
4. Justification: Since $f'$ changes from $+$ to $-$ at $x = -1$, $f(-1)$ is a relative max. Since $f'$ changes from $-$ to $+$ at $x = 3$, $f(3)$ is a relative min.

# Example 5: Finding Points of Inflection

Find the points of inflection for $f(x) = x^4 - 4x^3 + 2$.
Solution:

1. $f'(x) = 4x^3 - 12x^2$.
2. $f''(x) = 12x^2 - 24x = 12x(x - 2)$.
3. Possible POIs at $x = 0, 2$.
4. Sign Chart for $f''$:
   • $(-\infty, 0)$: $f''(-1) = (-)(-) = +$ (Concave Up)
   • $(0, 2)$: $f''(1) = (+)(-) = -$ (Concave Down)
   • $(2, \infty)$: $f''(3) = (+)(+) = +$ (Concave Up)
5. POIs at $x=0$ and $x=2$ because concavity changes.

# Example 6: Optimization (Maximum Area)

A farmer has 100 meters of fencing and wants to enclose a rectangular field adjacent to a straight river (no fence needed along the river). Find the dimensions for max area.

{
  "title": "Optimization: Fencing a Field by a River",
  "width": 520,
  "height": 300,
  "background": "grid",
  "elements": [
    { "type": "surface", "x1": 50, "y1": 60, "x2": 450, "y2": 60 },
    { "type": "label", "x": 250, "y": 45, "text": "RIVER (No Fence Needed)", "bold": true, "color": "#3b82f6" },
    { "type": "line", "x1": 150, "y1": 60, "x2": 150, "y2": 220, "color": "#6366f1", "width": 3 },
    { "type": "line", "x1": 150, "y1": 220, "x2": 350, "y2": 220, "color": "#6366f1", "width": 3 },
    { "type": "line", "x1": 350, "y1": 220, "x2": 350, "y2": 60, "color": "#6366f1", "width": 3 },
    { "type": "label", "x": 130, "y": 140, "text": "w", "bold": true },
    { "type": "label", "x": 370, "y": 140, "text": "w", "bold": true },
    { "type": "label", "x": 250, "y": 240, "text": "l", "bold": true },
    { "type": "label", "x": 250, "y": 140, "text": "Area = l * w", "color": "#10b981" }
  ]
}

Solution:

1. Variables: $w$ (width), $l$ (length parallel to river).
2. Constraint: $2w + l = 100 \implies l = 100 - 2w$.
3. Objective: $A = l \cdot w = (100 - 2w)w = 100w - 2w^2$.
4. Differentiate: $A'(w) = 100 - 4w$.
5. Solve: $100 - 4w = 0 \implies w = 25$.
6. $l = 100 - 2(25) = 50$.
7. Max Area: $25 \times 50 = 1250$ m$^2$.

# Example 7: Second Derivative Test

Classify critical points of $f(x) = x + \frac{4}{x}$ using the Second Derivative Test.
Solution:

1. $f'(x) = 1 - 4x^{-2} = 1 - \frac{4}{x^2}$.
2. $f'(x) = 0 \implies x^2 = 4 \implies x = \pm 2$.
3. $f''(x) = 8x^{-3} = \frac{8}{x^3}$.
4. Test $x=2$: $f''(2) = \frac{8}{8} = 1 > 0 \implies$ Rel Min at $x=2$.
5. Test $x=-2$: $f''(-2) = \frac{8}{-8} = -1 < 0 \implies$ Rel Max at $x=-2$.

# Example 8: Rolle's Theorem

Verify Rolle's Theorem for $f(x) = \sin(x)$ on $[0, \pi]$.
Solution:

1. $f(x)$ is continuous and differentiable.
2. $f(0) = \sin(0) = 0$.
3. $f(\pi) = \sin(\pi) = 0$.
4. Since $f(0) = f(\pi)$, there must be $c \in (0, \pi)$ such that $f'(c) = 0$.
5. $f'(x) = \cos(x)$. $\cos(c) = 0 \implies c = \pi/2$.

# Example 9: Interpreting $f'$ graph

If the graph of $f'$ is a downward-opening parabola with roots at $x=1$ and $x=5$, find the interval where $f$ is increasing.
Solution:

1. $f$ is increasing where $f'(x) > 0$.
2. For a downward parabola with roots at 1 and 5, the function is positive between the roots.
3. Interval: $(1, 5)$.

# Example 10: Connecting $f'$ and $f''$

If $f'(x) = e^x(x-2)$, find where $f$ is concave up.
Solution:

1. $f''(x) = e^x(x-2) + e^x(1) = e^x(x-2+1) = e^x(x-1)$.
2. $e^x$ is always positive. $f''(x) = 0$ at $x=1$.
3. $f''(x) > 0$ for $x > 1$.
4. $f$ is concave up on $(1, \infty)$.

# Example 11: MVT with a Table

# Example 12: Rational Function Extrema

Find the relative extrema of $f(x) = \frac{x^2}{x-1}$.
Solution:

1. $f'(x) = \frac{(x-1)(2x) - x^2(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2}$.
2. Critical points at $x = 0, 2$ ($x=1$ is not in domain).
3. Sign change at $x=0$: $f'$ goes from $+$ to $- \implies$ Rel Max.
4. Sign change at $x=2$: $f'$ goes from $-$ to $+ \implies$ Rel Min.

# Example 13: Absolute Min on Open Interval

Find the absolute minimum of $f(x) = x \ln x$ for $x > 0$.
Solution:

1. $f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
2. $\ln x + 1 = 0 \implies \ln x = -1 \implies x = e^{-1} = 1/e$.
3. $f''(x) = 1/x$. Since $1/x > 0$ for all $x > 0$, the function is always concave up.
4. Therefore, the relative minimum at $x = 1/e$ is the absolute minimum.

# Example 14: Optimization (Box Volume)

A square sheet of metal with side 12 inches has squares of side $x$ cut from corners to make an open box. Maximize volume.
Solution:

1. $V = x(12-2x)^2 = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x$.
2. $V' = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) = 12(x-6)(x-2)$.
3. Critical points: $x=2, 6$.
4. $x=6$ is impossible (cuts the whole sheet).
5. At $x=2$, $V = 2(8)^2 = 128$ in$^3$.

# Example 15: Concavity of Trig Function

Find intervals of concavity for $f(x) = x + \cos x$ on $[0, 2\pi]$.
Solution:

1. $f'(x) = 1 - \sin x$.
2. $f''(x) = -\cos x$.
3. $-\cos x = 0$ at $x = \pi/2, 3\pi/2$.
4. Sign of $f''$:
   • $(0, \pi/2)$: $-\cos(0.1) \approx -$ (Concave Down)
   • $(\pi/2, 3\pi/2)$: $-\cos(\pi) = +$ (Concave Up)
   • $(3\pi/2, 2\pi)$: $-\cos(6) = -$ (Concave Down)

# Example 16: Non-Differentiable Point

Find critical points for $f(x) = x^{2/3}$.
Solution:

1. $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
2. $f'(x)$ is never 0, but it is undefined at $x=0$.
3. Critical point: $x=0$. (This is a cusp).

# Example 17: MVT Failure

Why does MVT not apply to $f(x) = |x|$ on $[-1, 1]$?
Solution:
While $f(x)$ is continuous on $[-1, 1]$, it is not differentiable at $x=0$, which is inside the open interval $(-1, 1)$.

# Example 18: Optimization (Closest Point)

Find the point on $y = \sqrt{x}$ closest to $(4, 0)$.
Solution:

1. Distance $d = \sqrt{(x-4)^2 + (y-0)^2}$.
2. Minimize $d^2 = S = (x-4)^2 + (\sqrt{x})^2 = x^2 - 8x + 16 + x = x^2 - 7x + 16$.
3. $S' = 2x - 7 = 0 \implies x = 3.5$.
4. Point: $(3.5, \sqrt{3.5})$.

# Example 19: Second Derivative Test Inconclusive

Show that the 2nd Derivative test is inconclusive for $f(x) = x^4$ at $x=0$.
Solution:

1. $f'(x) = 4x^3 \implies f'(0) = 0$.
2. $f''(x) = 12x^2 \implies f''(0) = 0$.
3. Since $f''(0) = 0$, we cannot determine the nature of the point. (Using 1st derivative test, we see it's a minimum).

# Example 20: Sketching from $f'$ and $f''$

Sketch $f$ given: $f(0)=0, f'(x) > 0$ for all $x$, and $f''(x) < 0$ for $x < 0$ and $f''(x) > 0$ for $x > 0$.
Solution:
The graph always goes up. To the left of the y-axis, it is a "frown" shape. To the right, it is a "cup" shape. This is the graph of $f(x) = x^3$ shifted or similar.

# Example 21: Rectilinear Motion and Extrema

A particle's position is $s(t) = t^3 - 9t^2 + 24t$. Find the minimum velocity on $t \in [0, 5]$.
Solution:

1. Velocity $v(t) = 3t^2 - 18t + 24$.
2. To minimize velocity, find $v'(t) = a(t) = 6t - 18$.
3. Critical point for $v(t)$ is $t=3$.
4. Candidates: $v(0) = 24$, $v(3) = 27 - 54 + 24 = -3$, $v(5) = 75 - 90 + 24 = 9$.
5. Min velocity is $-3$ at $t=3$.

# Example 22: Identifying POI

If $f''(x) = (x-1)^2(x-5)$, find the x-coordinate of the point of inflection.
Solution:

1. $f''(x) = 0$ at $x=1$ and $x=5$.
2. Check sign changes:
   • Left of 1: $(1-1)^2 \dots$ actually check $0$: $(-1)^2(-5) = -$.
   • Between 1 and 5: check 2: $(1)^2(-3) = -$. (No sign change at 1).
   • Right of 5: check 6: $(5)^2(1) = +$.
3. Point of Inflection only at $x=5$.

# Example 23: Implicit Optimization

Find the maximum $y$ value for the circle $x^2 + y^2 = 25$.
Solution:

1. Differentiate implicitly: $2x + 2y y' = 0 \implies y' = -x/y$.
2. Max/Min when $y' = 0 \implies x = 0$.
3. Substitute $x=0$ into circle: $0^2 + y^2 = 25 \implies y = \pm 5$.
4. Max $y = 5$.

# Example 24: Revenue Maximization

Demand function $p = 100 - 0.01x$. Maximize Revenue $R = xp$.
Solution:

1. $R = x(100 - 0.01x) = 100x - 0.01x^2$.
2. $R' = 100 - 0.02x = 0 \implies x = 5000$.
3. $R'' = -0.02$ (always concave down, so it's a max).

# Example 25: Inflection Point of $e^{-x^2}$

Find the inflection points of the Gaussian $f(x) = e^{-x^2}$.
Solution:

1. $f'(x) = -2x e^{-x^2}$.
2. $f''(x) = -2e^{-x^2} + (-2x)(-2x e^{-x^2}) = e^{-x^2}(4x^2 - 2)$.
3. $4x^2 - 2 = 0 \implies x^2 = 1/2 \implies x = \pm 1/\sqrt{2}$.

# Example 26: Constant Function and MVT

For $f(x) = 5$ on $[0, 10]$, find $c$ for MVT.
Solution:
$f'(c) = \frac{5-5}{10} = 0$. Since $f'(x) = 0$ for all $x$, every $c \in (0, 10)$ satisfies the theorem.

# Example 27: Absolute Extrema on Infinite Interval

Find absolute extrema for $f(x) = \frac{x}{x^2+1}$.
Solution:

1. $\lim_{x \to \pm \infty} f(x) = 0$.
2. $f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$.
3. Critical points $x = \pm 1$.
4. $f(1) = 0.5$, $f(-1) = -0.5$.
5. Since $0.5 > 0$ and $-0.5 < 0$, Abs Max is $0.5$, Abs Min is $-0.5$.

# Example 28: Optimization (Minimum Material)

A cylindrical can must hold $16\pi$ cm$^3$. Minimize surface area.
Solution:

1. $V = \pi r^2 h = 16\pi \implies h = 16/r^2$.
2. $A = 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r(16/r^2) = 2\pi r^2 + 32\pi/r$.
3. $A' = 4\pi r - 32\pi/r^2 = 0 \implies 4\pi r^3 = 32\pi \implies r^3 = 8 \implies r = 2$.
4. $h = 16/4 = 4$.

# Example 29: Justification with $f''$

Given $f'(3)=0$ and $f''(x) = x^2-10$, classify $x=3$.
Solution:
$f''(3) = 3^2 - 10 = -1$.
Since $f'(3)=0$ and $f''(3) < 0$, there is a relative maximum at $x=3$ by the Second Derivative Test.

# Example 30: Vertical Tangent

Identify where $f(x) = \sqrt[3]{x-2}$ has a vertical tangent.
Solution:
$f'(x) = \frac{1}{3}(x-2)^{-2/3} = \frac{1}{3\sqrt[3]{(x-2)^2}}$.
The derivative is undefined at $x=2$. Since the limit of $f'$ as $x \to 2$ is $\infty$, there is a vertical tangent at $x=2$.

# Example 31: Monotonicity of $f(x) = x + \sin x$

Is $f(x) = x + \sin x$ ever decreasing?
Solution:
$f'(x) = 1 + \cos x$. Since $-1 \leq \cos x \leq 1$, $0 \leq f'(x) \leq 2$.
$f'(x)$ is never negative, so the function is never decreasing.

# Example 32: Number of Roots

Use Rolle's Theorem to show $f(x) = x^3 + x - 1$ has exactly one real root.
Solution:

1. $f(0) = -1$ and $f(1) = 1$. By Intermediate Value Theorem, there is at least one root.
2. Suppose there are two roots $a$ and $b$. Then $f(a)=f(b)=0$.
3. By Rolle's, $f'(c) = 0$ for some $c \in (a, b)$.
4. But $f'(x) = 3x^2 + 1$, which is always $\geq 1$.
5. Contradiction. Thus, only one root exists.

# Example 33: Connecting $f, f', f''$ through signs

If $f(x) > 0, f'(x) < 0,$ and $f''(x) > 0$ for all $x$, describe the graph.
Solution:
The graph is above the x-axis, moving downward, but "bending" upward (decreasing at a decreasing rate). It looks like an exponential decay curve $e^{-x}$.

# Example 34: Concavity and Tangent Line Approximation

If $f(2) = 5, f'(2) = 3$, and $f''(x) > 0$ for all $x$, is the tangent line approximation at $x=2.1$ an overestimate or underestimate?
Solution:
Since $f''(x) > 0$, the function is concave up. For concave up functions, the tangent line stays below the curve. Therefore, the approximation is an underestimate.

# Example 35: Extreme Value Theorem Logic

Does $f(x) = 1/x$ have an absolute maximum on $(0, 1)$?
Solution:
No. The interval $(0, 1)$ is open. The EVT requires a closed interval. As $x \to 0^+$, $f(x) \to \infty$, so there is no maximum.