Applications of Derivatives
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# Applications of Derivatives ## 🎯 Learning Objectives By the end of this topic, you will be able to: 1. **Find** critical points of a function 2.

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
If 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)=x3−3x2+1.
Step 1 — Critical points: f′(x)=3x2−6x=3x(x−2)=0⟹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)
Example 2: Find global extrema of f(x)=x2−4x+5 on [0,3].
Step 1 — Critical points: f′(x)=2x−4=0⟹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
Example 3: Find intervals of increase/decrease and concavity for f(x)=x3−6x2+9x+1.
Step 1 — Increase/Decrease: f′(x)=3x2−12x+9=3(x−1)(x−3)
f′(x)>0 on (−∞,1) and (3,∞) → increasing
f′(x)<0 on (1,3) → decreasing
Step 2 — Concavity: f′′(x)=6x−12=6(x−2)
f′′(x)>0 on (2,∞) → concave up
f′′(x)<0 on (−∞,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)=x3, f′(0)=0, but x=0 is an inflection point, not an extremum.
📝 Practice Questions
Q1: Find critical points of f(x)=2x3−9x2+12x+3.f′(x)=6x2−18x+12=6(x2−3x+2)=6(x−1)(x−2)=0 x=1,2x=1, x=2 Q2: Use the 2nd derivative test to classify critical points of f(x)=x3−3x.f′(x)=3x2−3=3(x−1)(x+1)→x=±1 f′′(x)=6x, f′′(1)=6>0 → local min, f′′(−1)=−6<0 → local maxx=1 (min),x=−1 (max) Q3: Find intervals of increase for f(x)=x4−4x3.f′(x)=4x3−12x2=4x2(x−3) Critical points: x=0,3 Test intervals: (−∞,0): f′(−1)=4(−1)2(−4)=−16<0 → decreasing (0,3): f′(1)=4(1)2(−2)=−8<0 → decreasing (3,∞): f′(4)=4(16)(1)=64>0 → increasingIncreasing on (3,∞) Q4: Find global max and min of f(x)=x2−2x on [0,2].f′(x)=2x−2=0→x=1 f(0)=0, f(1)=−1, f(2)=0 Max = 0 (at x=0 or x=2), Min = -1 (at x=1)Max 0, Min −1 Q5: Find inflection points of f(x)=x4−6x2.f′′(x)=12x2−12=12(x−1)(x+1)=0→x=±1 Check sign change: yes, concavity changes at both. f(1)=1−6=−5, f(−1)=1−6=−5(−1,−5), (1,−5) Q6: If f′(x)>0 for all x, what can you say about f?f is strictly increasing.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 ≥0). So inflection point, not extremum.No — inflection point Q8: Find the maximum area of a rectangle with perimeter 20.2l+2w=20→l+w=10→l=10−w A(w)=w(10−w)=10w−w2 A′(w)=10−2w=0→w=5, l=5 Max area = 5×5=2525
🔗 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