Applications of Derivatives — Monotonicity, Extrema, Concavity & Optimization

Layered Deep-Dive: This chapter reconstructs Applications of Derivatives from first principles. We master Fermat's Interior Extremum Theorem, identify critical points, perform First and Second Derivative Tests, classify concavity and inflection points, and solve closed-interval global optimization problems for IIT Madras Quiz 3.

1. Orientation & Intuition: Peaks, Troughs, and Bends

• When a roller-coaster is climbing, its slope is positive: $f'(x) > 0$.
• When it is descending, its slope is negative: $f'(x) < 0$.
• At the very peak of a hill or the bottom of a valley, the tangent line momentarily levels out completely: $f'(x) = 0$.

          Peak (Local Max) f'(c) = 0, f''(c) < 0
             .-'''-.
            /       \   Inflection Point f''(x) = 0
           /         \       .
          /           '._ _.' \
                               \
                                '---. Valley (Local Min) f'(d) = 0, f''(d) > 0

2. Terminology, Notation & Core Theorems

3. Fermat's Theorem & Critical Points

Fermat's Interior Extremum Theorem

If a function $f$ has a local extremum (maximum or minimum) at an interior point $c$, and $f'(c)$ exists, then:

$f'(c) = 0$

4. Classifying Extrema: The Two Tests

4.1 The First Derivative Test

4.2 The Second Derivative Test

• If $f''(c) > 0$: the curve is concave up ($\cup$) $\implies$ Local Minimum at $c$.
• If $f''(c) < 0$: the curve is concave down ($\cap$) $\implies$ Local Maximum at $c$.
• If $f''(c) = 0$: the test is inconclusive! (You must revert to the First Derivative Test).

5. Global Extrema on a Closed Interval $[a, b]$

The Closed-Interval Algorithm:

1. Differentiate $f(x)$ to find $f'(x)$.
2. Find all critical points $c_1, c_2, \dots$ in the open interval $(a, b)$ where $f'(c) = 0$ or $f'(c)$ does not exist. (Discard any points outside $(a, b)$).
3. Evaluate $f(x)$ at:
   • All interior critical points: $f(c_1), f(c_2), \dots$
   • Both boundary endpoints: $f(a)$ and $f(b)$.
4. The largest resulting value is the Absolute Maximum, and the smallest is the Absolute Minimum.

6. Worked Exam Dissections

Dissection 1: Comprehensive Curve Analysis

1. Determine the intervals where $f$ is strictly increasing and decreasing.
2. Find and classify all local extrema.
3. Determine intervals of concavity and locate all points of inflection.

• On $(-\infty, 1)$: choose $x = 0 \implies f'(0) = 6(-1)(-2) = +12 > 0 \implies$ Increasing.
• On $(1, 2)$: choose $x = 1.5 \implies f'(1.5) = 6(0.5)(-0.5) = -1.5 < 0 \implies$ Decreasing.
• On $(2, \infty)$: choose $x = 3 \implies f'(3) = 6(2)(1) = +12 > 0 \implies$ Increasing.

• At $x = 1$: $f'$ changes from $+$ to $-$ $\implies$ Local Maximum at $(1, f(1))$. $f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 3 = 2 - 9 + 12 - 3 = 2$
• At $x = 2$: $f'$ changes from $-$ to $+$ $\implies$ Local Minimum at $(2, f(2))$. $f(2) = 2(8) - 9(4) + 12(2) - 3 = 16 - 36 + 24 - 3 = 1$

• For $x < 1.5$: $f''(x) < 0 \implies$ Concave Down.
• For $x > 1.5$: $f''(x) > 0 \implies$ Concave Up. Since $f''$ changes sign across $x = 1.5$, this is a true point of inflection: $f(1.5) = 2(3.375) - 9(2.25) + 12(1.5) - 3 = 6.75 - 20.25 + 18 - 3 = 1.5$ $\text{Point of Inflection: } (1.5, 1.5)$

Dissection 2: Absolute Extrema on a Closed Interval

• $x = 0 \in (-1, 3)$ $\implies$ Keep.
• $x = 2 \in (-1, 3)$ $\implies$ Keep.
• $x = -2 \notin (-1, 3)$ $\implies$ Discard! (Falls outside the given interval).

1. Interior critical point $x = 0$: $f(0) = 0^4 - 8(0)^2 + 5 = 5$
2. Interior critical point $x = 2$: $f(2) = 2^4 - 8(2)^2 + 5 = 16 - 32 + 5 = -11$
3. Left endpoint $x = -1$: $f(-1) = (-1)^4 - 8(-1)^2 + 5 = 1 - 8 + 5 = -2$
4. Right endpoint $x = 3$: $f(3) = 3^4 - 8(3)^2 + 5 = 81 - 72 + 5 = 14$

• Largest value: $14$ (at $x = 3$).
• Smallest value: $-11$ (at $x = 2$).

7. Common Traps & Pitfalls in IITM Calculus Exams

1. Forgetting to Test Endpoints: In closed-interval optimization, the global maximum or minimum often occurs at an endpoint $a$ or $b$ where $f'(x) \neq 0$. Never declare an extremum without evaluating the boundary points.
2. Missing Critical Points Where $f'$ is Undefined: For functions like $f(x) = |x|$ or $f(x) = x^{2/3}$, the derivative at $x = 0$ does not exist. $x = 0$ is still a valid critical point and in fact the location of an absolute minimum.
3. Assuming $f''(c) = 0$ Implies an Inflection Point: Just because $f''(c) = 0$ does not guarantee an inflection point. For $f(x) = x^4$, $f''(0) = 12(0)^2 = 0$, but $f''(x) = 12x^2 \ge 0$ everywhere, so the concavity never changes sign. $x=0$ is a local minimum, not an inflection point!

8. Practice & Quiz Questions

Question 1 (MCQ)

• (A) $x = 0$
• (B) $x = 1$
• (C) $x = -1$
• (D) $x = e$

Question 2 (MSQ)

• (A) If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at $c$.
• (B) If $f$ has a local minimum at $c$, then $f''(c)$ must be strictly positive.
• (C) If $f''(x) > 0$ for all $x \in (a, b)$, then the tangent line to $f$ at any point in $(a, b)$ lies entirely below the graph of $f$.
• (D) A point $c$ where $f''(c) = 0$ is always a point of inflection.

• (A) True: The Second Derivative Test confirms local maximum.
• (B) False: For $f(x) = x^4$, $x = 0$ is a local minimum, but $f''(0) = 0$, which is not strictly positive.
• (C) True: This is the geometric definition of strict convex/concave up curvature.
• (D) False: For $f(x) = x^4$, $f''(0) = 0$, but concavity does not change sign (concave up on both sides).

9. Active Recall Flashcards

1. What is the formal definition of a critical point of $f(x)$? Answer: A point $c$ in the domain of $f$ where either $f'(c) = 0$ or $f'(c)$ is undefined.
2. What does the Second Derivative Test conclude if $f'(c) = 0$ and $f''(c) = 0$? Answer: The test is inconclusive; you must use the First Derivative Test.
3. What is an inflection point? Answer: A point on the curve where the concavity strictly changes sign (from concave up to concave down, or vice versa).
4. How do you find global extrema on a closed interval $[a, b]$? Answer: Evaluate $f(x)$ at all critical points inside $(a, b)$ and at both endpoints $a$ and $b$; the largest is the global maximum and the smallest is the global minimum.