Quiz 2

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

📋 Prerequisites


📖 Core Content

14.1 Critical Points

A critical point of ff is a point cc in the domain where f(c)=0f'(c) = 0 (horizontal tangent) or f(c)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)f'(x) sign on intervalff behavior
f(x)>0f'(x) > 0ff is increasing
f(x)<0f'(x) < 0ff is decreasing
f(x)=0f'(x) = 0Possible extremum

14.3 First Derivative Test

At a critical point cc:
Sign change of ff' at ccConclusion
++ to -Local maximum at cc
- to ++Local minimum at cc
No sign changeNeither (saddle/inflection)

14.4 Second Derivative and Concavity

f(x)f''(x) signConcavity
f(x)>0f''(x) > 0Concave up (U-shaped)
f(x)<0f''(x) < 0Concave down (∩-shaped)
Inflection point: Where concavity changes (f(x)=0f''(x) = 0 or DNE).

14.5 Second Derivative Test

If cc is a critical point with f(c)=0f'(c) = 0:
f(c)f''(c)Conclusion
f(c)>0f''(c) > 0Local minimum
f(c)<0f''(c) < 0Local maximum
f(c)=0f''(c) = 0Test inconclusive

14.6 Global Extrema on [a,b][a, b]

Recipe:
  1. Find all critical points in (a,b)(a, b)
  2. Evaluate ff at critical points and at endpoints a,ba, b
  3. Largest value = global max; smallest = global min

14.7 Worked Examples

Example 1: Find local extrema of f(x)=x33x2+1f(x) = x^3 - 3x^2 + 1. Step 1 — Critical points: f(x)=3x26x=3x(x2)=0    x=0,2f'(x) = 3x^2 - 6x = 3x(x-2) = 0 \implies x = 0, 2 Step 2 — 2nd Derivative Test: f(x)=6x6f''(x) = 6x - 6 f(0)=6<0f''(0) = -6 < 0 → local max at (0,1)(0, 1) f(2)=6>0f''(2) = 6 > 0 → local min at (2,3)(2, -3)
Local max at (0,1), 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)=x24x+5f(x) = x^2 - 4x + 5 on [0,3][0, 3]. Step 1 — Critical points: f(x)=2x4=0    x=2f'(x) = 2x - 4 = 0 \implies x = 2 Step 2 — Evaluate: f(0)=5f(0) = 5, f(2)=48+5=1f(2) = 4 - 8 + 5 = 1, f(3)=912+5=2f(3) = 9 - 12 + 5 = 2 Step 3 — Compare: Max = 5 at x=0x=0, Min = 1 at x=2x=2
Global max 5 at x=0, Global 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)=x36x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1. Step 1 — Increase/Decrease: f(x)=3x212x+9=3(x1)(x3)f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3) f(x)>0f'(x) > 0 on (,1)(-\infty, 1) and (3,)(3, \infty) → increasing f(x)<0f'(x) < 0 on (1,3)(1, 3) → decreasing Step 2 — Concavity: f(x)=6x12=6(x2)f''(x) = 6x - 12 = 6(x-2) f(x)>0f''(x) > 0 on (2,)(2, \infty) → concave up f(x)<0f''(x) < 0 on (,2)(-\infty, 2) → concave down Inflection point at x=2x = 2 (f(2)=824+18+1=3f(2) = 8 - 24 + 18 + 1 = 3)

📐 Key Formulas — Summary Table

ConceptFormula/TestWhen to Use
Critical pointf(c)=0f'(c) = 0 or DNECandidates for extrema
Increasingf(x)>0f'(x) > 0
Decreasingf(x)<0f'(x) < 0
1st Derivative TestSign change of ff' at ccClassify critical points
2nd Derivative Testf(c)>0f''(c) > 0 (min), f(c)<0f''(c) < 0 (max)When ff'' exists
Concave upf(x)>0f''(x) > 0
Concave downf(x)<0f''(x) < 0
Inflection pointf(x)=0f''(x) = 0 , concavity changes
Global max on [a,b][a,b]Largest ff value at critical pt or endpointOptimization

⚠️ 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][a,b] might occur at endpoints, even if no critical point gives an extremum there.

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

f(c)=0f'(c)=0 is necessary but not sufficient. Example: f(x)=x3f(x) = x^3, f(0)=0f'(0) = 0, but x=0x=0 is an inflection point, not an extremum.

📝 Practice Questions

Q1: Find critical points of f(x)=2x39x2+12x+3f(x) = 2x^3 - 9x^2 + 12x + 3.
f(x)=6x218x+12=6(x23x+2)=6(x1)(x2)=0f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2) = 0 x=1,2x = 1, 2
x=1, x=2\boxed{x=1,\ x=2} Q2: Use the 2nd derivative test to classify critical points of f(x)=x33xf(x) = x^3 - 3x.
f(x)=3x23=3(x1)(x+1)x=±1f'(x) = 3x^2 - 3 = 3(x-1)(x+1) \to x = \pm 1 f(x)=6xf''(x) = 6x, f(1)=6>0f''(1) = 6 > 0 → local min, f(1)=6<0f''(-1) = -6 < 0 → local max
x=1 (min),x=1 (max)\boxed{x=1 \text{ (min)}, x=-1 \text{ (max)}} Q3: Find intervals of increase for f(x)=x44x3f(x) = x^4 - 4x^3.
f(x)=4x312x2=4x2(x3)f'(x) = 4x^3 - 12x^2 = 4x^2(x-3) Critical points: x=0,3x=0, 3 Test intervals: (,0)(-\infty,0): f(1)=4(1)2(4)=16<0f'(-1) = 4(-1)^2(-4) = -16 < 0 → decreasing (0,3)(0,3): f(1)=4(1)2(2)=8<0f'(1) = 4(1)^2(-2) = -8 < 0 → decreasing (3,)(3,\infty): f(4)=4(16)(1)=64>0f'(4) = 4(16)(1) = 64 > 0 → increasing
Increasing on (3,)\boxed{\text{Increasing on } (3,\infty)} Q4: Find global max and min of f(x)=x22xf(x) = x^2 - 2x on [0,2][0, 2].
f(x)=2x2=0x=1f'(x) = 2x - 2 = 0 \to x = 1 f(0)=0f(0) = 0, f(1)=1f(1) = -1, f(2)=0f(2) = 0 Max = 0 (at x=0x=0 or x=2x=2), Min = -1 (at x=1x=1)
Max 0, Min 1\boxed{\text{Max } 0,\ \text{Min } -1} Q5: Find inflection points of f(x)=x46x2f(x) = x^4 - 6x^2.
f(x)=12x212=12(x1)(x+1)=0x=±1f''(x) = 12x^2 - 12 = 12(x-1)(x+1) = 0 \to x = \pm 1 Check sign change: yes, concavity changes at both. f(1)=16=5f(1) = 1 - 6 = -5, f(1)=16=5f(-1) = 1 - 6 = -5
(1,5), (1,5)\boxed{(-1,-5),\ (1,-5)} Q6: If f(x)>0f'(x) > 0 for all xx, what can you say about ff?
ff is strictly increasing.
Strictly increasing\boxed{\text{Strictly increasing}} Q7: For f(x)=(x1)3f(x) = (x-1)^3, does ff have a local extremum at x=1x=1?
f(x)=3(x1)2f'(x) = 3(x-1)^2, f(1)=0f'(1) = 0 but f(x)f'(x) doesn't change sign (always 0\geq 0). So inflection point, not extremum.
No — inflection point\boxed{\text{No — inflection point}} Q8: Find the maximum area of a rectangle with perimeter 20.
2l+2w=20l+w=10l=10w2l + 2w = 20 \to l + w = 10 \to l = 10-w A(w)=w(10w)=10ww2A(w) = w(10-w) = 10w - w^2 A(w)=102w=0w=5A'(w) = 10 - 2w = 0 \to w = 5, l=5l = 5 Max area = 5×5=255 \times 5 = 25
25\boxed{25}

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