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Home/Calculus/Chapter 4/Applications

CALC-4.5

4-5 hours

Applications of Derivatives

Apply derivatives to analyze function behavior: monotonicity, extrema, convexity, inflection points, and asymptotes. Learn to sketch graphs and prove important inequalities.

Learning Objectives

Use derivatives to determine monotonicity of functions

Find and classify local and global extrema

Determine convexity using the second derivative

Find inflection points where concavity changes

Identify vertical, horizontal, and oblique asymptotes

Apply the first and second derivative tests

Sketch graphs using derivative information

Prove inequalities using convexity (Jensen, Young, Hölder)

1. Monotonicity and Extrema

Theorem 4.18: Monotonicity Criterion

Let $f \in C[a,b] \cap D(a,b)$ . Then:

$f'(x) > 0$ on $(a,b)$ ⟹ $f$ is strictly increasing on $[a,b]$
$f'(x) < 0$ on $(a,b)$ ⟹ $f$ is strictly decreasing on $[a,b]$
$f'(x) = 0$ on $(a,b)$ ⟹ $f$ is constant on $[a,b]$

Theorem 4.19: First Derivative Test

Suppose $f'(x_0) = 0$ . Then:

$f'$ changes from + to − at $x_0$ ⟹ local maximum at $x_0$
$f'$ changes from − to + at $x_0$ ⟹ local minimum at $x_0$
$f'$ does not change sign ⟹ neither max nor min

Theorem 4.20: Second Derivative Test

If $f'(x_0) = 0$ and $f''(x_0)$ exists:

$f''(x_0) < 0$ ⟹ local maximum at $x_0$
$f''(x_0) > 0$ ⟹ local minimum at $x_0$
$f''(x_0) = 0$ ⟹ test is inconclusive

Proof of Theorem 4.18 (Monotonicity Criterion):

Suppose $f'(x) > 0$ on $(a, b)$ . For $x_1 < x_2$ in $[a, b]$ :

By MVT, $\exists \xi \in (x_1, x_2)$ :

f(x_2) - f(x_1) = f'(\xi)(x_2 - x_1)

Since $f'(\xi) > 0$ and $x_2 - x_1 > 0$ , we have $f(x_2) > f(x_1)$ .

∎

Example 4.18: Finding Monotonicity Intervals

Problem: Find the intervals of monotonicity for $f(x) = x^3 - 3x^2 + 2$ .

Solution:

Step 1: Find $f'(x) = 3x^2 - 6x = 3x(x - 2)$ .

Step 2: Critical points: $x = 0$ and $x = 2$ .

Step 3: Sign analysis:

• $x < 0$ : $f'(x) > 0$ (increasing)
• $0 < x < 2$ : $f'(x) < 0$ (decreasing)
• $x > 2$ : $f'(x) > 0$ (increasing)

Local max at $x = 0$ : $f(0) = 2$ . Local min at $x = 2$ : $f(2) = -2$ .

Example 4.18b: Using Second Derivative Test

Problem: Classify the critical points of $f(x) = x^4 - 4x^3$ .

Solution:

$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$ .

Critical points: $x = 0$ and $x = 3$ .

$f''(x) = 12x^2 - 24x = 12x(x - 2)$ .

• At $x = 0$ : $f''(0) = 0$ (inconclusive)
• At $x = 3$ : $f''(3) = 12 \cdot 3 \cdot 1 = 36 > 0$ (local minimum)

For $x = 0$ , use first derivative test: $f' < 0$ on both sides, so neither max nor min.

Remark 4.1: Global vs Local Extrema

To find global extrema on a closed interval $[a, b]$ :

Find all critical points (where $f' = 0$ or undefined)
Evaluate $f$ at all critical points and endpoints
Compare all values to find global max/min

Note: On an open interval or unbounded domain, global extrema may not exist!

Example 4.18c: Global Extrema on Closed Interval

Problem: Find the global max and min of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$ .

Solution:

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ . Critical points: $x = \pm 1$ .

Evaluate at critical points and endpoints:

$f(-2) = -8 + 6 + 1 = -1$
$f(-1) = -1 + 3 + 1 = 3$
$f(1) = 1 - 3 + 1 = -1$
$f(2) = 8 - 6 + 1 = 3$

Global max: 3 at $x = -1$ and $x = 2$ . Global min: -1 at $x = -2$ and $x = 1$ .

Theorem 4.20b: Higher-Order Derivative Test

If $f'(x_0) = f''(x_0) = \cdots = f^{(n-1)}(x_0) = 0$ but $f^{(n)}(x_0) \neq 0$ :

If $n$ is even and $f^{(n)}(x_0) > 0$ : local minimum
If $n$ is even and $f^{(n)}(x_0) < 0$ : local maximum
If $n$ is odd: inflection point (neither max nor min)

Example 4.18d: Optimization Word Problem

Problem: A farmer has 100m of fencing. What dimensions give maximum rectangular area?

Solution:

Let length = $x$ , width = $y$ . Constraint: $2x + 2y = 100$ , so $y = 50 - x$ .

Area: $A(x) = x(50 - x) = 50x - x^2$ for $x \in [0, 50]$ .

$A'(x) = 50 - 2x = 0$ gives $x = 25$ .

$A''(x) = -2 < 0$ , so $x = 25$ is a maximum.

Answer: Square with side 25m gives max area = 625 m².

Remark 4.1b: Critical Points Summary

Types of Critical Points:

Type 1: $f'(x_0) = 0$ (stationary points)
Type 2: $f'(x_0)$ doesn't exist but $f(x_0)$ exists (corner points)

Both types can be local extrema. Example: $|x|$ has a minimum at $x = 0$ where $f'$ doesn't exist.

Example 4.18e: Function with Corner Point

Problem: Find the extrema of $f(x) = |x^2 - 1|$ .

Solution:

$f(x) = |x^2 - 1| = \begin{cases} x^2 - 1 & |x| \geq 1 \\ 1 - x^2 & |x| < 1 \end{cases}$

Critical points: $x = 0$ (stationary), $x = \pm 1$ (corners).

Values: $f(0) = 1$ , $f(\pm 1) = 0$ .

Local max at $x = 0$ . Local min at $x = \pm 1$ .

Example 4.18f: Proving Inequality Using Derivatives

Problem: Prove that $e^x \geq 1 + x$ for all $x \in \mathbb{R}$ .

Solution:

Let $f(x) = e^x - 1 - x$ . Then $f'(x) = e^x - 1$ .

$f'(x) = 0$ at $x = 0$ . $f''(x) = e^x > 0$ , so $x = 0$ is a global min.

$f(0) = 1 - 1 - 0 = 0$ .

Since $f(x) \geq f(0) = 0$ , we have $e^x \geq 1 + x$ . ✓

2. Convexity

Definition 4.5: Convex Function

$f$ is convex on $[a,b]$ if for all $x_1, x_2 \in [a,b]$ and $t \in [0,1]$ :

f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)

Geometrically: the graph lies below any chord.

Theorem 4.21: Second Derivative Test for Convexity

Let $f \in C[a,b] \cap D^2(a,b)$ . Then:

$f''(x) \geq 0$ on $(a,b)$ ⟺ $f$ is convex on $[a,b]$
$f''(x) \leq 0$ on $(a,b)$ ⟺ $f$ is concave on $[a,b]$

Theorem 4.22: Jensen's Inequality

If $f$ is convex on $[a,b]$ , $x_i \in [a,b]$ , and $\sum t_i = 1$ with $t_i > 0$ :

f\left(\sum_{i=1}^n t_i x_i\right) \leq \sum_{i=1}^n t_i f(x_i)

Proof of Theorem 4.21 (Convexity Test):

Suppose $f'' \geq 0$ . For any $x_1 < x_2$ and $t \in (0, 1)$ , let $x = tx_1 + (1-t)x_2$ .

By Taylor with Lagrange remainder around $x$ :

f(x_1) = f(x) + f'(x)(x_1 - x) + \frac{f''(\xi_1)}{2}(x_1 - x)^2

f(x_2) = f(x) + f'(x)(x_2 - x) + \frac{f''(\xi_2)}{2}(x_2 - x)^2

Since $f'' \geq 0$ , both remainder terms are $\geq 0$ .

Computing $tf(x_1) + (1-t)f(x_2)$ and using $t(x_1 - x) + (1-t)(x_2 - x) = 0$ :

tf(x_1) + (1-t)f(x_2) \geq f(x) = f(tx_1 + (1-t)x_2)

∎

Remark 4.2: Geometric Interpretation of Convexity

Convex (concave up)

• Graph lies below chords
• Graph lies above tangent lines
• $f'' \geq 0$
• "Holds water"

Concave (concave down)

• Graph lies above chords
• Graph lies below tangent lines
• $f'' \leq 0$
• "Spills water"

Example 4.19: Testing Convexity

Problem: Determine the convexity of $f(x) = x^3 - 3x$ .

Solution:

$f'(x) = 3x^2 - 3$ , $f''(x) = 6x$ .

• $x < 0$ : $f''(x) < 0$ (concave)
• $x > 0$ : $f''(x) > 0$ (convex)

Inflection point at $x = 0$ .

Example 4.19b: Proving AM-GM Using Jensen

Problem: Prove $\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}$ for positive $x_i$ .

Solution:

Let $f(x) = -\ln x$ (convex since $f''(x) = 1/x^2 > 0$ ).

By Jensen with $t_i = 1/n$ :

-\ln\left(\frac{x_1 + \cdots + x_n}{n}\right) \leq \frac{1}{n}\sum_{i=1}^n (-\ln x_i)

\ln\left(\frac{x_1 + \cdots + x_n}{n}\right) \geq \frac{1}{n}\ln(x_1 \cdots x_n) = \ln\sqrt[n]{x_1 \cdots x_n}

Exponentiating: $\text{AM} \geq \text{GM}$ .

Corollary 4.8: Common Convex/Concave Functions

Convex Functions

• $e^x$
• $x^n$ for $n \geq 2$ , $x \geq 0$
• $|x|^p$ for $p \geq 1$
• $-\ln x$ for $x > 0$

Concave Functions

• $\ln x$ for $x > 0$
• $\sqrt{x}$ for $x \geq 0$
• $\sin x$ on $[0, \pi]$
• $x^p$ for $0 < p < 1$ , $x > 0$

Example 4.19c: Young's Inequality

Result: For $a, b > 0$ and $\frac{1}{p} + \frac{1}{q} = 1$ ( $p, q > 1$ ):

ab \leq \frac{a^p}{p} + \frac{b^q}{q}

This follows from the convexity of $e^x$ applied to $\frac{1}{p}\ln a^p + \frac{1}{q}\ln b^q$ .

3. Inflection Points

Definition 4.6: Inflection Point

A point $(x_0, f(x_0))$ is an inflection point if $f$ changes concavity at $x_0$ (from convex to concave or vice versa).

Theorem 4.23: Inflection Point Criteria

Necessary condition: If $(x_0, f(x_0))$ is an inflection point and $f''(x_0)$ exists, then $f''(x_0) = 0$ .

Sufficient condition: If $f''(x_0) = 0$ and $f'''(x_0) \neq 0$ , then $(x_0, f(x_0))$ is an inflection point.

Example 4.20: Finding Inflection Points

Problem: Find all inflection points of $f(x) = x^4 - 6x^2 + 8x + 1$ .

Solution:

$f'(x) = 4x^3 - 12x + 8$

$f''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$

$f''(x) = 0$ at $x = \pm 1$ .

Sign analysis of $f''$ :

• $x < -1$ : $f'' > 0$ (convex)
• $-1 < x < 1$ : $f'' < 0$ (concave)
• $x > 1$ : $f'' > 0$ (convex)

Inflection points at $(-1, f(-1))$ and $(1, f(1))$ .

Example 4.20b: When f'' = 0 is NOT an Inflection

Problem: Check if $f(x) = x^4$ has an inflection point at $x = 0$ .

Solution:

$f''(x) = 12x^2$ , so $f''(0) = 0$ .

But $f''(x) \geq 0$ for all $x$ — no sign change!

$x = 0$ is NOT an inflection point; it's a local minimum.

Remark 4.3b: Inflection Point Summary

To find inflection points:

Find where $f''(x) = 0$ or $f''$ is undefined
Check if $f''$ changes sign at each candidate
Calculate $f(x_0)$ to get the full point $(x_0, f(x_0))$

4. Asymptotes

Definition 4.7: Types of Asymptotes

Vertical: $x = a$ if $\lim_{x \to a} f(x) = \pm\infty$
Horizontal: $y = L$ if $\lim_{x \to \pm\infty} f(x) = L$
Oblique: $y = kx + b$ if $\lim_{x \to \pm\infty} [f(x) - (kx + b)] = 0$

Remark 4.3: Finding Oblique Asymptotes

For oblique asymptote $y = kx + b$ :

k = \lim_{x \to \infty} \frac{f(x)}{x}, \quad b = \lim_{x \to \infty} [f(x) - kx]

Example 4.17: Graphing with All Tools

Sketch $f(x) = \frac{x^3 - 1}{x^2}$ .

Domain: $x \neq 0$
Vertical asymptote: $x = 0$
Oblique asymptote: $k = \lim \frac{x^3-1}{x^3} = 1$ , $b = \lim (x - 1/x^2 - x) = 0$ , so $y = x$
Derivative: $f'(x) = 1 + 2/x^3$
Critical point: $x = -\sqrt[3]{2}$

Example 4.21: Finding All Asymptotes

Problem: Find all asymptotes of $f(x) = \frac{x^2 - 1}{x - 2}$ .

Solution:

Vertical: $\lim_{x \to 2^\pm} f(x) = \pm\infty$ , so $x = 2$ is vertical asymptote.

Oblique: $k = \lim_{x \to \infty} \frac{x^2 - 1}{x(x-2)} = \lim \frac{x^2}{x^2} = 1$

$b = \lim_{x \to \infty} \left(\frac{x^2-1}{x-2} - x\right) = \lim \frac{x^2 - 1 - x(x-2)}{x-2} = \lim \frac{2x - 1}{x - 2} = 2$

Vertical: $x = 2$ . Oblique: $y = x + 2$ .

Remark 4.4: Asymptote Decision Tree

Vertical asymptotes: Check where denominator = 0
Horizontal asymptotes: Compute $\lim_{x \to \pm\infty} f(x)$
If no horizontal asymptote, check for oblique: Compute $k = \lim f(x)/x$ and $b = \lim (f(x) - kx)$

Example 4.21b: Rational Function Graphing

Problem: Complete graph analysis for $f(x) = \frac{x^2}{x^2 - 4}$ .

Solution:

Domain: $x \neq \pm 2$

Symmetry: $f(-x) = f(x)$ (even function)

Intercepts: $(0, 0)$

Vertical asymptotes: $x = \pm 2$

Horizontal asymptote: $\lim_{x \to \infty} \frac{x^2}{x^2 - 4} = 1$ , so $y = 1$

Monotonicity: $f'(x) = \frac{-8x}{(x^2-4)^2}$

• $x < 0$ : $f' > 0$ (increasing)
• $x > 0$ : $f' < 0$ (decreasing)
• Local max at $x = 0$ : $f(0) = 0$

5. Complete Curve Sketching

Remark 4.5: Curve Sketching Checklist

Steps for Complete Analysis:

Domain: Where is $f$ defined?
Symmetry: Even ( $f(-x) = f(x)$ )? Odd ( $f(-x) = -f(x)$ )?
Intercepts: Where does graph cross axes?
Asymptotes: Vertical, horizontal, oblique?
First derivative: Critical points, monotonicity, local extrema
Second derivative: Convexity, inflection points
Sketch: Combine all information

Example 4.22: Complete Analysis: e^(-x²)

Problem: Sketch $f(x) = e^{-x^2}$ (Gaussian function).

Solution:

Domain: $\mathbb{R}$
Symmetry: Even (symmetric about y-axis)
Range: $(0, 1]$
Horizontal asymptote: $y = 0$ as $x \to \pm\infty$
First derivative: $f'(x) = -2xe^{-x^2}$
- • $f'(x) = 0$ at $x = 0$
- • $f' > 0$ for $x < 0$ , $f' < 0$ for $x > 0$
- • Global max at $x = 0$ : $f(0) = 1$
Second derivative: $f''(x) = (4x^2 - 2)e^{-x^2}$
- • $f''(x) = 0$ at $x = \pm 1/\sqrt{2}$
- • Inflection points at $x = \pm 1/\sqrt{2}$

Example 4.22b: Complete Analysis: x ln x

Problem: Sketch $f(x) = x \ln x$ for $x > 0$ .

Solution:

Domain: $(0, \infty)$
Behavior at 0: $\lim_{x \to 0^+} x \ln x = 0$
First derivative: $f'(x) = \ln x + 1$
- • $f'(x) = 0$ at $x = 1/e$
- • Local min at $x = 1/e$ : $f(1/e) = -1/e$
Second derivative: $f''(x) = 1/x > 0$ for all $x > 0$ (convex)

Example 4.22c: Complete Analysis: xe^(-x)

Problem: Sketch $f(x) = xe^{-x}$ .

Solution:

Domain: $\mathbb{R}$
Intercepts: $(0, 0)$
Limits: $\lim_{x \to -\infty} = -\infty$ , $\lim_{x \to \infty} = 0$
First derivative: $f'(x) = e^{-x}(1 - x)$
- • $f'(x) = 0$ at $x = 1$
- • Global max at $x = 1$ : $f(1) = 1/e$
Second derivative: $f''(x) = e^{-x}(x - 2)$
- • Inflection point at $x = 2$

Example 4.22d: Rational Function Analysis

Problem: Sketch $f(x) = \frac{x}{x^2 + 1}$ .

Solution:

Domain: $\mathbb{R}$
Symmetry: Odd function ( $f(-x) = -f(x)$ )
Intercepts: $(0, 0)$
Horizontal asymptote: $y = 0$
First derivative: $f'(x) = \frac{1 - x^2}{(x^2+1)^2}$
- • Critical points: $x = \pm 1$
- • Local max at $x = 1$ : $f(1) = 1/2$
- • Local min at $x = -1$ : $f(-1) = -1/2$

Remark 4.5b: Signs Table Template

How to organize analysis:

Interval	$f'(x)$	$f''(x)$	Behavior
$(a, c_1)$	+	+	↗ convex
$(c_1, c_2)$	−	+	↘ convex
...	...	...	...

6. Optimization Problems

Example 4.23: Box with Maximum Volume

Problem: An open box is made from a 12×12 square by cutting squares of side $x$ from corners. Find $x$ for maximum volume.

Solution:

Volume: $V(x) = x(12 - 2x)^2$ for $x \in [0, 6]$ .

$V(x) = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x$

$V'(x) = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) = 12(x-2)(x-6)$

Critical points: $x = 2, 6$ . Since $V(0) = V(6) = 0$ :

Answer: $x = 2$ gives max volume $V(2) = 2 \cdot 64 = 128$ cubic units.

Example 4.23b: Minimum Distance

Problem: Find the point on $y = x^2$ closest to $(0, 1)$ .

Solution:

Distance squared: $D(x) = x^2 + (x^2 - 1)^2$

$D'(x) = 2x + 2(x^2 - 1)(2x) = 2x(1 + 2x^2 - 2) = 2x(2x^2 - 1)$

$D'(x) = 0$ at $x = 0$ or $x = \pm 1/\sqrt{2}$ .

Comparing: $D(0) = 1$ , $D(\pm 1/\sqrt{2}) = 1/2 + 1/4 = 3/4$ .

Answer: Closest points are $(\pm 1/\sqrt{2}, 1/2)$ .

Example 4.23c: Economics: Maximizing Profit

Problem: Revenue $R(q) = 100q - q^2$ , Cost $C(q) = 20 + 10q$ . Find quantity for max profit.

Solution:

Profit: $P(q) = R(q) - C(q) = 100q - q^2 - 20 - 10q = -q^2 + 90q - 20$

$P'(q) = -2q + 90 = 0$ gives $q = 45$ .

$P''(q) = -2 < 0$ , so this is a maximum.

Answer: Max profit at $q = 45$ units: $P(45) = 2005$ .

Example 4.23d: Cylinder with Maximum Volume

Problem: A cylinder has surface area 100 cm². Find dimensions for maximum volume.

Solution:

Surface: $2\pi r^2 + 2\pi rh = 100$ , so $h = \frac{100 - 2\pi r^2}{2\pi r} = \frac{50}{\pi r} - r$ .

Volume: $V = \pi r^2 h = \pi r^2 \left(\frac{50}{\pi r} - r\right) = 50r - \pi r^3$

$V'(r) = 50 - 3\pi r^2 = 0$ gives $r = \sqrt{\frac{50}{3\pi}}$ .

Answer: $r = \sqrt{\frac{50}{3\pi}} \approx 2.3$ cm, $h = 2r \approx 4.6$ cm.

Remark 4.6: Optimization Strategy

General Approach:

Draw a diagram and label variables
Identify the quantity to optimize (objective function)
Find constraints relating variables
Express objective in terms of one variable
Find domain of the resulting function
Apply calculus: find critical points, test for max/min
Check endpoints if domain is closed
Verify the answer makes sense

Example 4.23e: Least Material for a Can

Problem: Design a cylindrical can holding 1000 cm³ using minimum material.

Solution:

Volume: $V = \pi r^2 h = 1000$ , so $h = \frac{1000}{\pi r^2}$ .

Surface: $S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \frac{2000}{r}$

$S'(r) = 4\pi r - \frac{2000}{r^2} = 0$ gives $r^3 = \frac{500}{\pi}$ .

$r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42$ cm.

Answer: Optimal when $h = 2r$ (height = diameter).

Example 4.23f: Related Rates Application

Problem: Water fills a conical tank (r = 3m, h = 6m at top) at 2 m³/min. How fast is water level rising when h = 3m?

Solution:

Similar triangles: $\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .

Volume: $V = \frac{1}{3}\pi r^2 h = \frac{\pi h^3}{12}$

Differentiate: $\frac{dV}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$

At $h = 3$ : $2 = \frac{9\pi}{4} \cdot \frac{dh}{dt}$

Answer: $\frac{dh}{dt} = \frac{8}{9\pi} \approx 0.28$ m/min.

7. Important Inequalities via Convexity

Theorem 4.24: Hölder's Inequality

For $p, q > 1$ with $\frac{1}{p} + \frac{1}{q} = 1$ :

\sum_{i=1}^n |a_i b_i| \leq \left(\sum_{i=1}^n |a_i|^p\right)^{1/p} \left(\sum_{i=1}^n |b_i|^q\right)^{1/q}

Special case: $p = q = 2$ gives Cauchy-Schwarz.

Corollary 4.9: Cauchy-Schwarz Inequality

\left(\sum_{i=1}^n a_i b_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right)

Example 4.24: Proving Cauchy-Schwarz

Method: Use the convexity of $f(x) = x^2$ .

Consider $P(t) = \sum (a_i + t b_i)^2 \geq 0$ for all $t$ .

Expanding: $P(t) = At^2 + 2Bt + C \geq 0$ where

$A = \sum b_i^2, \quad B = \sum a_i b_i, \quad C = \sum a_i^2$

Discriminant $\leq 0$ : $4B^2 - 4AC \leq 0$ , so $B^2 \leq AC$ .

Remark 4.7: Applications of These Inequalities

In Analysis

• Proving convergence of series
• Bounding integrals
• L^p space theory

In Applications

• Machine learning (regularization)
• Signal processing
• Economics (utility functions)

Example 4.24b: Using AM-GM in Optimization

Problem: Find the minimum of $f(x) = x + \frac{4}{x}$ for $x > 0$ .

Solution (Two Methods):

Method 1 (Calculus): $f'(x) = 1 - 4/x^2 = 0$ gives $x = 2$ . Min = 4.

Method 2 (AM-GM): $\frac{x + 4/x}{2} \geq \sqrt{x \cdot 4/x} = 2$

So $x + 4/x \geq 4$ , equality when $x = 4/x$ , i.e., $x = 2$ .

Example 4.24c: Power Mean Inequality

Result: For positive $a_i$ and $r < s$ :

\left(\frac{a_1^r + \cdots + a_n^r}{n}\right)^{1/r} \leq \left(\frac{a_1^s + \cdots + a_n^s}{n}\right)^{1/s}

Special cases: $r \to 0$ gives GM, $r = 1$ gives AM, $r = 2$ gives QM.

Corollary 4.10: QM-AM-GM-HM Chain

\sqrt{\frac{\sum a_i^2}{n}} \geq \frac{\sum a_i}{n} \geq \sqrt[n]{\prod a_i} \geq \frac{n}{\sum 1/a_i}

Quadratic Mean ≥ Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean

Remark 4.8: Historical Note

Jensen's inequality (1906) is named after Danish mathematician Johan Jensen. It provides a unified framework for proving many classical inequalities including AM-GM, Cauchy-Schwarz, and Hölder's inequality, all through the concept of convexity.

Example 4.24d: Minkowski's Inequality

Result: For $p \geq 1$ :

\left(\sum_{i=1}^n |a_i + b_i|^p\right)^{1/p} \leq \left(\sum_{i=1}^n |a_i|^p\right)^{1/p} + \left(\sum_{i=1}^n |b_i|^p\right)^{1/p}

This is the triangle inequality in $L^p$ space. For $p = 2$ , it gives the familiar $\|a + b\| \leq \|a\| + \|b\|$ .

Remark 4.9: Summary: Key Inequalities

Inequality	Statement
AM-GM	$\frac{a+b}{2} \geq \sqrt{ab}$
Cauchy-Schwarz	$(\sum a_i b_i)^2 \leq (\sum a_i^2)(\sum b_i^2)$
Young	$ab \leq \frac{a^p}{p} + \frac{b^q}{q}$
Jensen	$f(\mathbb{E}[X]) \leq \mathbb{E}[f(X)]$ (convex $f$ )

Historical Context

The systematic study of optimization using calculus began with Fermat and Newton in the 17th century. The theory of convex functions was developed by Jensen, Minkowski, and others in the early 20th century, becoming fundamental to modern optimization, economics, and machine learning.

Applications Practice

Questions

Correct

Accuracy

f'(x) > 0

for all

x \in (a, b)

, then

f

is:

Easy

Not attempted

f''(x) > 0

(a, b)

, then

f

is:

Easy

Not attempted

At an inflection point, we must have:

Medium

Not attempted

For

f(x) = x^3

, classify

x = 0

Medium

Not attempted

\lim_{x \to \infty} f(x) = L

(finite), then

y = L

is:

Easy

Not attempted

The second derivative test says: if

f'(x_0) = 0

and

f''(x_0) < 0

, then:

Easy

Not attempted

For oblique asymptote

y = kx + b

, how do we find

k

Medium

Not attempted

Jensen's inequality for convex

f

: if

\sum t_i = 1

(

t_i > 0

), then:

Hard

Not attempted

For

f(x) = x^3 - 3x

, the local maximum occurs at:

Medium

Not attempted

A function is concave on an interval if:

Easy

Not attempted

The AM-GM inequality can be proved using Jensen's inequality with:

Hard

Not attempted

For curve sketching, the first step should be:

Easy

Not attempted

f'(x_0) = 0

f''(x_0) = 0

, and

f'''(x_0) \neq 0

, then

x_0

is:

Medium

Not attempted

To maximize area with fixed perimeter, the optimal shape is:

Medium

Not attempted

Frequently Asked Questions

How do I determine if a critical point is max, min, or neither?

Use the first derivative test (check sign change of f') or second derivative test (check sign of f''). If f'(x₀)=0 and f''(x₀)=0, use higher derivatives or first derivative test.

What's the difference between local and global extrema?

Local extrema compare values in a neighborhood; global extrema are the absolute max/min over the entire domain. A local extremum may not be global.

How do inflection points relate to convexity?

Inflection points are where the function changes from convex to concave (or vice versa). At such points, f'' = 0 and f'' changes sign.

When should I use asymptotes in graphing?

Always check for vertical asymptotes (where f → ±∞), horizontal asymptotes (behavior at ±∞), and oblique asymptotes (when f(x)/x → k ≠ 0 as x → ∞).

What's the difference between convex and concave?

Convex (concave up): f'' ≥ 0, graph holds water, lies below chords. Concave (concave down): f'' ≤ 0, graph spills water, lies above chords.

How do I apply Jensen's inequality?

For convex f: f(weighted average) ≤ weighted average of f. For concave f: reverse the inequality. The weights must sum to 1.

When is the second derivative test inconclusive?

When f'(x₀) = 0 and f''(x₀) = 0. Then use the first derivative test or check higher derivatives.

How do I find oblique asymptotes?

Compute k = lim(x→∞) f(x)/x and b = lim(x→∞) [f(x) - kx]. If both limits exist and k ≠ 0, y = kx + b is an oblique asymptote.

Can a function have both horizontal and oblique asymptotes?

Not in the same direction. At each of ±∞, a function can have at most one asymptote (horizontal or oblique).

How do optimization problems work?

1) Identify the quantity to optimize. 2) Express it as a function of one variable. 3) Find critical points. 4) Use derivative tests to classify. 5) Check boundary conditions.

Key Takeaways

Monotonicity

$f' > 0 \Rightarrow$ increasing; $f' < 0 \Rightarrow$ decreasing

Convexity

$f'' > 0 \Rightarrow$ convex; $f'' < 0 \Rightarrow$ concave

Extrema Tests

First derivative: sign change. Second derivative: sign at critical point.

Inflection Points

$f'' = 0$ and $f''$ changes sign

Study Tips

Sign Charts

Always make a sign chart for $f'$ and $f''$ . Mark critical points and test intervals.

Check Endpoints

For global extrema on closed intervals, always evaluate at endpoints too.

Optimization Strategy

Express everything in one variable, find the domain, then apply calculus.

Asymptotes First

When graphing, find asymptotes early — they guide the overall shape.

Common Mistakes to Avoid

✗
Assuming f'(x₀) = 0 is sufficient for extremum
It's necessary, not sufficient. Check sign change or second derivative.
✗
Confusing f''(x₀) = 0 with inflection point
f'' must also change sign at an inflection point.
✗
Forgetting to check domain boundaries
Global extrema may occur at endpoints, not just critical points.
✗
Mixing up convex and concave
Remember: convex = "holds water" = f'' > 0.

Chapter 4 Complete!

Congratulations! You've mastered the fundamentals of differentiation:

Definition of Derivative: Limits, differentiability, basic rules
Chain Rule: Composite functions, implicit differentiation
Mean Value Theorems: Rolle, Lagrange, Cauchy
L'Hospital & Taylor: Limits, series expansions
Applications: Extrema, convexity, graphing, optimization

You can now continue to Chapter 5: Integration (coming soon).