Maths I — Week 5: Applications of Derivatives

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Maths I — Week 5: Applications of Derivatives

The Big Idea: The derivative tells you the slope at any point. But slope is direction. A positive slope means the function is climbing. A zero slope means the function has peaked or bottomed out. This is how we find the best (and worst) values of any mathematical model.

1. Monotonicity — Is the Function Rising or Falling?

Condition	Meaning
$f'(x) > 0$ on $(a, b)$	$f$ is strictly increasing — moving up-right
$f'(x) < 0$ on $(a, b)$	$f$ is strictly decreasing — moving down-right
$f'(x) = 0$ at $x = c$	Stationary (critical) point — function momentarily "flat"

If two functions

f

(strictly increasing) and

g

(strictly decreasing) intersect at

x_0

, then

f(x) \geq g(x)

for all

x \geq x_0

, and

g(x) \geq f(x)

for all

x \leq x_0

. This is directly tested in Week 7, Question 2.

2. Finding Critical Points

A critical point is where the derivative is zero or undefined.

Procedure

Step 1: Compute $f'(x)$ .
Step 2: Set $f'(x) = 0$ and solve for $x$ .
Step 3: Check additional points where $f'(x)$ is undefined (e.g., at sharp corners or domain boundaries).

Example: For

f(x) = x^3 - 3x^2 + 2

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ .
Critical points: $x = 0$ and $x = 2$ .

3. First Derivative Test

After finding critical points, check the sign of $f'$ on either side:

Sign Change	Conclusion
$f'$ changes $+$ to $-$	Local Maximum
$f'$ changes $-$ to $+$	Local Minimum
No sign change	Inflection Point (neither max nor min)

Continuing the example:

At $x = 0$ : $f'$ changes from $+$ to $-$ → Local Max, $f(0) = 2$ .
At $x = 2$ : $f'$ changes from $-$ to $+$ → Local Min, $f(2) = -2$ .

4. Second Derivative Test (Faster Route)

At a critical point

c

where

f'(c) = 0

Condition	Conclusion
$f''(c) < 0$	Local Maximum (concave down — like ∩)
$f''(c) > 0$	Local Minimum (concave up — like ∪)
$f''(c) = 0$	Inconclusive — use the First Derivative Test

Remember it visually: concave

down = peak (like an upside-down bowl) = maximum. Concave up = valley (like a right-side-up bowl) = minimum.

5. Concavity and Inflection Points

Concavity describes the direction the curve is "bending":

$f''(x) > 0$ : Concave Up (slopes are increasing).
$f''(x) < 0$ : Concave Down (slopes are decreasing).

An Inflection Point is where concavity flips (changes sign). It is not necessarily a critical point.

6. Global vs. Local Extrema

Local: Only better than neighbors in a small neighborhood.
Global: Best (or worst) over the entire domain.

Rule for closed intervals $[a, b]$ :

Always evaluate $f$ at: all interior critical points AND the endpoints $f(a)$ and $f(b)$ . The true global max/min is the largest/smallest among all these values.

7. Pattern A — The Optimization Protocol

What to recognize: A word problem asking to maximize or minimize something (area, profit, time, cost).

Abstract Solution (Strategy)

[Define objective]: Identify the quantity to maximize/minimize. Write it as a function.
[Apply constraint]: Use any given restrictions to reduce to one variable.
[Differentiate]: $\frac{d}{dx}(\text{objective}) = 0$ .
[Verify]: Use the second derivative test or boundary check.

Procedure

Step 1: Let the variable be $x$ .
Step 2: Write the objective function $P(x)$ .
Step 3: Compute $P'(x)$ and set to 0.
Step 4: Confirm it is a max/min using $P''(x)$ .

Worked Example:

Find two positive numbers whose sum is 20 and whose product is maximum.

Let numbers be $x$ and $20 - x$ . Product $P = x(20-x)$ .
$P'(x) = 20 - 2x = 0 \implies x = 10$ .
$P''(x) = -2 < 0$ → Maximum.
Result: Both numbers are 10, product = 100.

8. Pattern B — Profit Maximization (Week 8 Link)

Question: Price

p(x) = 1000 - x

. Cost

C(x) = 30000 + 300x

for

x \leq 400

. Maximize profit for

x \leq 400

Abstract Solution (Strategy)

[Revenue]: $R(x) = x \cdot p(x) = x(1000 - x)$ .
[Profit]: $P(x) = R(x) - C(x)$ .
[Maximize]: Vertex of the resulting quadratic.

Procedure

Step 1: $P(x) = (1000x - x^2) - (30000 + 300x) = -x^2 + 700x - 30000$ .
Step 2: $P'(x) = -2x + 700 = 0 \implies x = 350$ .
Step 3: $350 \leq 400$ ✓.
Result: Produce 350 units.

Profit =

Total Revenue (not unit price!) minus Total Cost. Forgetting to multiply price by quantity is the most common error in these problems.

9. Tangent and Normal Lines

Tangent at $(a, f(a))$ :

y - f(a) = f'(a)(x - a)

Normal at $(a, f(a))$ :

y - f(a) = -\frac{1}{f'(a)}(x - a)

Worked Example:

Find the tangent and normal to $y = \sqrt{x}$ at $x = 4$ .

$f(4) = 2$ . Point: $(4, 2)$ .
$f'(x) = \frac{1}{2\sqrt{x}} \implies f'(4) = \frac{1}{4}$ .
Tangent: $y - 2 = \frac{1}{4}(x-4) \implies y = \frac{x}{4} + 1$ .
Normal: $y - 2 = -4(x-4) \implies y = -4x + 18$ .

10. Common Mistakes

Mistake	Why it happens	Correct approach
Not checking endpoints for global extrema	Focusing only on critical points in the interior.	On a closed interval, all endpoints must be evaluated alongside critical points.
Misidentifying inflection as extremum	Seeing $f'(c) = 0$ and assuming it's a max/min.	Always check the sign change of $f'$ (First Derivative Test). No sign change = inflection.
Incorrect profit formula	Using price as revenue.	Revenue = Price × Quantity. $R(x) = x \cdot p(x)$ .
Second Derivative Test on inconclusive points	$f''(c) = 0$ is not "neutral" — it means the test fails entirely.	Fall back to the First Derivative Test in this case.

11. Flashcards

<Flashcard front="If f'(a) = 0 and f''(a) > 0, what do we conclude?" back="f has a local MINIMUM at x = a. (Concave up = valley = minimum.)" /> <Flashcard front="What is the slope of the normal line if f'(a) = 3?" back="-1/3. The normal slope is always the negative reciprocal of the tangent slope." /> <Flashcard front="What does f'(x) > 0 mean?" back="The function is strictly increasing on that interval." /> <Flashcard front="For global extrema on [a,b], what must you always check?" back="All interior critical points PLUS both endpoints f(a) and f(b). The largest/smallest is the global answer." /> <Flashcard front="Sign change test: f' goes from + to -, what is at that critical point?" back="A local MAXIMUM." />

12. Practice Targets

Analyze $f(x) = x^4 - 4x^3$ — find all critical points, classify them, identify inflection points.
Optimization: A farmer has 100m of fence and wants to enclose a rectangular field maximum area (one side is a wall). Find the optimal dimensions.
Attempt Questions 19 and 20 from Graded Assignment 8 (direct optimization).
Given two monotone functions that intersect, predict the relative ordering for different $x$ ranges.

13. Connections

Connects to	How
Week 4 — Derivatives	All optimization requires computing $f'(x)$ first.
Week 8 — Profit Maximization	The exact optimization pattern is tested in the Week 8 graded assignment.
Week 7 — Monotone Functions	Understanding strictly increasing/decreasing functions is tested directly in Week 7, Question 2.

Maths I — Week 5: Applications of Derivatives

1. Monotonicity — Is the Function Rising or Falling?

2. Finding Critical Points

Procedure

3. First Derivative Test

4. Second Derivative Test (Faster Route)

5. Concavity and Inflection Points

6. Global vs. Local Extrema

7. Pattern A — The Optimization Protocol

Abstract Solution (Strategy)

Procedure

8. Pattern B — Profit Maximization (Week 8 Link)

Abstract Solution (Strategy)

Procedure

9. Tangent and Normal Lines

10. Common Mistakes

11. Flashcards

12. Practice Targets

13. Connections

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