Applications of Derivatives

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# Applications of Derivatives ## ๐ŸŽฏ Learning Objectives By the end of this topic, you will be able to: 1. **Find** critical points of a function 2.

Applications of Derivatives

๐ŸŽฏ 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

๐Ÿ“‹ Prerequisites


๐Ÿ“– Core Content

14.1 Critical Points

A critical point of ff is a point cc in the domain where fโ€ฒ(c)=0f'(c) = 0 (horizontal tangent) or fโ€ฒ(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

fโ€ฒ(x)f'(x) sign on intervalff behavior
fโ€ฒ(x)>0f'(x) > 0ff is increasing
fโ€ฒ(x)<0f'(x) < 0ff is decreasing
fโ€ฒ(x)=0f'(x) = 0Possible extremum

14.3 First Derivative Test

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

14.4 Second Derivative and Concavity

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

14.5 Second Derivative Test

If cc is a critical point with fโ€ฒ(c)=0f'(c) = 0:
fโ€ฒโ€ฒ(c)f''(c)Conclusion
fโ€ฒโ€ฒ(c)>0f''(c) > 0Local minimum
fโ€ฒโ€ฒ(c)<0f''(c) < 0Local maximum
fโ€ฒโ€ฒ(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)=x3โˆ’3x2+1f(x) = x^3 - 3x^2 + 1. Step 1 โ€” Critical points: fโ€ฒ(x)=3x2โˆ’6x=3x(xโˆ’2)=0โ€…โ€ŠโŸนโ€…โ€Šx=0,2f'(x) = 3x^2 - 6x = 3x(x-2) = 0 \implies x = 0, 2 Step 2 โ€” 2nd Derivative Test: fโ€ฒโ€ฒ(x)=6xโˆ’6f''(x) = 6x - 6 fโ€ฒโ€ฒ(0)=โˆ’6<0f''(0) = -6 < 0 โ†’ local max at (0,1)(0, 1) fโ€ฒโ€ฒ(2)=6>0f''(2) = 6 > 0 โ†’ local min at (2,โˆ’3)(2, -3)
Localย maxย atย (0,1),ย localย minย atย (2,โˆ’3)\boxed{\text{Local max at } (0,1),\ \text{local min at } (2,-3)}
Example 2: Find global extrema of f(x)=x2โˆ’4x+5f(x) = x^2 - 4x + 5 on [0,3][0, 3]. Step 1 โ€” Critical points: fโ€ฒ(x)=2xโˆ’4=0โ€…โ€ŠโŸนโ€…โ€Šx=2f'(x) = 2x - 4 = 0 \implies x = 2 Step 2 โ€” Evaluate: f(0)=5f(0) = 5, f(2)=4โˆ’8+5=1f(2) = 4 - 8 + 5 = 1, f(3)=9โˆ’12+5=2f(3) = 9 - 12 + 5 = 2 Step 3 โ€” Compare: Max = 5 at x=0x=0, Min = 1 at x=2x=2
Globalย maxย 5ย atย x=0,ย Globalย minย 1ย atย 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)=x3โˆ’6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1. Step 1 โ€” Increase/Decrease: fโ€ฒ(x)=3x2โˆ’12x+9=3(xโˆ’1)(xโˆ’3)f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3) fโ€ฒ(x)>0f'(x) > 0 on (โˆ’โˆž,1)(-\infty, 1) and (3,โˆž)(3, \infty) โ†’ increasing fโ€ฒ(x)<0f'(x) < 0 on (1,3)(1, 3) โ†’ decreasing Step 2 โ€” Concavity: fโ€ฒโ€ฒ(x)=6xโˆ’12=6(xโˆ’2)f''(x) = 6x - 12 = 6(x-2) fโ€ฒโ€ฒ(x)>0f''(x) > 0 on (2,โˆž)(2, \infty) โ†’ concave up fโ€ฒโ€ฒ(x)<0f''(x) < 0 on (โˆ’โˆž,2)(-\infty, 2) โ†’ concave down Inflection point at x=2x = 2 (f(2)=8โˆ’24+18+1=3f(2) = 8 - 24 + 18 + 1 = 3)

๐Ÿ“ Key Formulas โ€” Summary Table

ConceptFormula/TestWhen to Use
Critical pointfโ€ฒ(c)=0f'(c) = 0 or DNECandidates for extrema
Increasingfโ€ฒ(x)>0f'(x) > 0
Decreasingfโ€ฒ(x)<0f'(x) < 0
1st Derivative TestSign change of fโ€ฒf' at ccClassify critical points
2nd Derivative Testfโ€ฒโ€ฒ(c)>0f''(c) > 0 (min), fโ€ฒโ€ฒ(c)<0f''(c) < 0 (max)When fโ€ฒโ€ฒf'' exists
Concave upfโ€ฒโ€ฒ(x)>0f''(x) > 0
Concave downfโ€ฒโ€ฒ(x)<0f''(x) < 0
Inflection pointfโ€ฒโ€ฒ(x)=0f''(x) = 0 , concavity changes
Global max on [a,b][a,b]Largest ff value at critical pt or endpointOptimization

โš ๏ธ 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 fโ€ฒ(c)=0f'(c)=0 Implies Extremum

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

๐Ÿ“ Practice Questions

Q1: Find critical points of f(x)=2x3โˆ’9x2+12x+3f(x) = 2x^3 - 9x^2 + 12x + 3.
fโ€ฒ(x)=6x2โˆ’18x+12=6(x2โˆ’3x+2)=6(xโˆ’1)(xโˆ’2)=0f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2) = 0 x=1,2x = 1, 2
x=1,ย x=2\boxed{x=1,\ x=2} Q2: Use the 2nd derivative test to classify critical points of f(x)=x3โˆ’3xf(x) = x^3 - 3x.
fโ€ฒ(x)=3x2โˆ’3=3(xโˆ’1)(x+1)โ†’x=ยฑ1f'(x) = 3x^2 - 3 = 3(x-1)(x+1) \to x = \pm 1 fโ€ฒโ€ฒ(x)=6xf''(x) = 6x, fโ€ฒโ€ฒ(1)=6>0f''(1) = 6 > 0 โ†’ local min, fโ€ฒโ€ฒ(โˆ’1)=โˆ’6<0f''(-1) = -6 < 0 โ†’ local max
x=1ย (min),x=โˆ’1ย (max)\boxed{x=1 \text{ (min)}, x=-1 \text{ (max)}} Q3: Find intervals of increase for f(x)=x4โˆ’4x3f(x) = x^4 - 4x^3.
fโ€ฒ(x)=4x3โˆ’12x2=4x2(xโˆ’3)f'(x) = 4x^3 - 12x^2 = 4x^2(x-3) Critical points: x=0,3x=0, 3 Test intervals: (โˆ’โˆž,0)(-\infty,0): fโ€ฒ(โˆ’1)=4(โˆ’1)2(โˆ’4)=โˆ’16<0f'(-1) = 4(-1)^2(-4) = -16 < 0 โ†’ decreasing (0,3)(0,3): fโ€ฒ(1)=4(1)2(โˆ’2)=โˆ’8<0f'(1) = 4(1)^2(-2) = -8 < 0 โ†’ decreasing (3,โˆž)(3,\infty): fโ€ฒ(4)=4(16)(1)=64>0f'(4) = 4(16)(1) = 64 > 0 โ†’ increasing
Increasingย onย (3,โˆž)\boxed{\text{Increasing on } (3,\infty)} Q4: Find global max and min of f(x)=x2โˆ’2xf(x) = x^2 - 2x on [0,2][0, 2].
fโ€ฒ(x)=2xโˆ’2=0โ†’x=1f'(x) = 2x - 2 = 0 \to x = 1 f(0)=0f(0) = 0, f(1)=โˆ’1f(1) = -1, f(2)=0f(2) = 0 Max = 0 (at x=0x=0 or x=2x=2), Min = -1 (at x=1x=1)
Maxย 0,ย Minย โˆ’1\boxed{\text{Max } 0,\ \text{Min } -1} Q5: Find inflection points of f(x)=x4โˆ’6x2f(x) = x^4 - 6x^2.
fโ€ฒโ€ฒ(x)=12x2โˆ’12=12(xโˆ’1)(x+1)=0โ†’x=ยฑ1f''(x) = 12x^2 - 12 = 12(x-1)(x+1) = 0 \to x = \pm 1 Check sign change: yes, concavity changes at both. f(1)=1โˆ’6=โˆ’5f(1) = 1 - 6 = -5, f(โˆ’1)=1โˆ’6=โˆ’5f(-1) = 1 - 6 = -5
(โˆ’1,โˆ’5),ย (1,โˆ’5)\boxed{(-1,-5),\ (1,-5)} Q6: If fโ€ฒ(x)>0f'(x) > 0 for all xx, what can you say about ff?
ff is strictly increasing.
Strictlyย increasing\boxed{\text{Strictly increasing}} Q7: For f(x)=(xโˆ’1)3f(x) = (x-1)^3, does ff have a local extremum at x=1x=1?
fโ€ฒ(x)=3(xโˆ’1)2f'(x) = 3(x-1)^2, fโ€ฒ(1)=0f'(1) = 0 but fโ€ฒ(x)f'(x) doesn't change sign (always โ‰ฅ0\geq 0). So inflection point, not extremum.
Noย โ€”ย inflectionย point\boxed{\text{No โ€” inflection point}} Q8: Find the maximum area of a rectangle with perimeter 20.
2l+2w=20โ†’l+w=10โ†’l=10โˆ’w2l + 2w = 20 \to l + w = 10 \to l = 10-w A(w)=w(10โˆ’w)=10wโˆ’w2A(w) = w(10-w) = 10w - w^2 Aโ€ฒ(w)=10โˆ’2w=0โ†’w=5A'(w) = 10 - 2w = 0 \to w = 5, l=5l = 5 Max area = 5ร—5=255 \times 5 = 25
25\boxed{25}

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