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Applications of Derivatives

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Applications of Derivatives

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Now · 🎯 Learning Objectives

Applications of Derivatives

🎯 Learning Objectives

By the end of this topic, you will be able to:

Find critical points of a function
Classify critical points using the first and second derivative tests
Determine intervals of increase/decrease and concavity
Find global maxima and minima on closed intervals
Sketch curves using derivative information

📋 Prerequisites

Derivatives — Definition & Rules — differentiation techniques
Polynomials — factoring, solving equations

📖 Core Content

14.1 Critical Points

A critical point of

f

is a point

c

in the domain where

f'(c) = 0

(horizontal tangent) or

f'(c)

does not exist (corner, cusp, vertical tangent). Why? Local maxima and minima can ONLY occur at critical points (but not every critical point is an extremum).

14.2 Increasing/Decreasing Test

$f'(x)$ sign on interval	$f$ behavior
$f'(x) > 0$	$f$ is increasing
$f'(x) < 0$	$f$ is decreasing
$f'(x) = 0$	Possible extremum

14.3 First Derivative Test

At a critical point

c

Sign change of $f'$ at $c$	Conclusion
$+$ to $-$	Local maximum at $c$
$-$ to $+$	Local minimum at $c$
No sign change	Neither (saddle/inflection)

14.4 Second Derivative and Concavity

$f''(x)$ sign	Concavity
$f''(x) > 0$	Concave up (U-shaped)
$f''(x) < 0$	Concave down (∩-shaped)

Inflection point: Where concavity changes (

f''(x) = 0

or DNE).

14.5 Second Derivative Test

c

is a critical point with

f'(c) = 0

$f''(c)$	Conclusion
$f''(c) > 0$	Local minimum
$f''(c) < 0$	Local maximum
$f''(c) = 0$	Test inconclusive

14.6 Global Extrema on $[a, b]$

Recipe:

Find all critical points in $(a, b)$
Evaluate $f$ at critical points and at endpoints $a, b$
Largest value = global max; smallest = global min

14.7 Worked Examples

Example 1: Find local extrema of

f(x) = x^3 - 3x^2 + 1

. Step 1 — Critical points:

f'(x) = 3x^2 - 6x = 3x(x-2) = 0 \implies x = 0, 2

Step 2 — 2nd Derivative Test:

f''(x) = 6x - 6

f''(0) = -6 < 0

→ local max at

(0, 1)

f''(2) = 6 > 0

→ local min at

(2, -3)

\boxed{\text{Local max at } (0,1),\ \text{local min at } (2,-3)}

Example 2: Find global extrema of

f(x) = x^2 - 4x + 5

[0, 3]

. Step 1 — Critical points:

f'(x) = 2x - 4 = 0 \implies x = 2

Step 2 — Evaluate:

f(0) = 5

f(2) = 4 - 8 + 5 = 1

f(3) = 9 - 12 + 5 = 2

Step 3 — Compare: Max = 5 at

x=0

, Min = 1 at

x=2

\boxed{\text{Global max } 5 \text{ at } x=0,\ \text{Global min } 1 \text{ at } x=2}

Example 3: Find intervals of increase/decrease and concavity for

f(x) = x^3 - 6x^2 + 9x + 1

. Step 1 — Increase/Decrease:

f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)

f'(x) > 0

(-\infty, 1)

and

(3, \infty)

→ increasing

f'(x) < 0

(1, 3)

→ decreasing Step 2 — Concavity:

f''(x) = 6x - 12 = 6(x-2)

f''(x) > 0

(2, \infty)

→ concave up

f''(x) < 0

(-\infty, 2)

→ concave down Inflection point at

x = 2

(

f(2) = 8 - 24 + 18 + 1 = 3

)

📐 Key Formulas — Summary Table

Concept	Formula/Test	When to Use
Critical point	$f'(c) = 0$ or DNE	Candidates for extrema
Increasing	$f'(x) > 0$
Decreasing	$f'(x) < 0$
1st Derivative Test	Sign change of $f'$ at $c$	Classify critical points
2nd Derivative Test	$f''(c) > 0$ (min), $f''(c) < 0$ (max)	When $f''$ exists
Concave up	$f''(x) > 0$
Concave down	$f''(x) < 0$
Inflection point	$f''(x) = 0$ , concavity changes
Global max on $[a,b]$	Largest $f$ value at critical pt or endpoint	Optimization

⚠️ Common Pitfalls

Pitfall 1: Confusing Local and Global Extrema

A local maximum is the highest point in its neighborhood. The global maximum is the highest overall. A function can have many local maxima but only one global maximum.

Pitfall 2: Forgetting Endpoints in Global Extrema

Global extrema on

[a,b]

might occur at endpoints, even if no critical point gives an extremum there.

Pitfall 3: Thinking $f'(c)=0$ Implies Extremum

f'(c)=0

is necessary but not sufficient. Example:

f(x) = x^3

f'(0) = 0

, but

x=0

is an inflection point, not an extremum.

📝 Practice Questions

Q1: Find critical points of $f(x) = 2x^3 - 9x^2 + 12x + 3$ .

$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2) = 0$ $x = 1, 2$

$\boxed{x=1,\ x=2}$ Q2: Use the 2nd derivative test to classify critical points of $f(x) = x^3 - 3x$ .

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1) \to x = \pm 1$ $f''(x) = 6x$ , $f''(1) = 6 > 0$ → local min, $f''(-1) = -6 < 0$ → local max

$\boxed{x=1 \text{ (min)}, x=-1 \text{ (max)}}$ Q3: Find intervals of increase for $f(x) = x^4 - 4x^3$ .

$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$ Critical points: $x=0, 3$ Test intervals: $(-\infty,0)$ : $f'(-1) = 4(-1)^2(-4) = -16 < 0$ → decreasing $(0,3)$ : $f'(1) = 4(1)^2(-2) = -8 < 0$ → decreasing $(3,\infty)$ : $f'(4) = 4(16)(1) = 64 > 0$ → increasing

$\boxed{\text{Increasing on } (3,\infty)}$ Q4: Find global max and min of $f(x) = x^2 - 2x$ on $[0, 2]$ .

$f'(x) = 2x - 2 = 0 \to x = 1$ $f(0) = 0$ , $f(1) = -1$ , $f(2) = 0$ Max = 0 (at $x=0$ or $x=2$ ), Min = -1 (at $x=1$ )

$\boxed{\text{Max } 0,\ \text{Min } -1}$ Q5: Find inflection points of $f(x) = x^4 - 6x^2$ .

$f''(x) = 12x^2 - 12 = 12(x-1)(x+1) = 0 \to x = \pm 1$ Check sign change: yes, concavity changes at both. $f(1) = 1 - 6 = -5$ , $f(-1) = 1 - 6 = -5$

$\boxed{(-1,-5),\ (1,-5)}$ Q6: If $f'(x) > 0$ for all $x$ , what can you say about $f$ ?

$f$ is strictly increasing.

$\boxed{\text{Strictly increasing}}$ Q7: For $f(x) = (x-1)^3$ , does $f$ have a local extremum at $x=1$ ?

$f'(x) = 3(x-1)^2$ , $f'(1) = 0$ but $f'(x)$ doesn't change sign (always $\geq 0$ ). So inflection point, not extremum.

$\boxed{\text{No — inflection point}}$ Q8: Find the maximum area of a rectangle with perimeter 20.

$2l + 2w = 20 \to l + w = 10 \to l = 10-w$ $A(w) = w(10-w) = 10w - w^2$ $A'(w) = 10 - 2w = 0 \to w = 5$ , $l = 5$ Max area = $5 \times 5 = 25$

$\boxed{25}$

🔗 Cross-References

Previous: Derivatives Definition & Rules
Next topic: Integration
Across courses: BSMA1003 Maths 2 (multivariable optimization); ML (gradient descent) Join Discord Previous8.1 Derivatives — Definition & RulesNext9.1 Indefinite Integrals