Applied mathematics - 2nd homework

INTUITIVE DEFINITION:
As x approaches a value, what does f(x) approach?

MATHEMATICAL NOTATION:
lim[x→a] f(x) = L

Meaning: f(x) gets arbitrarily close to L as x gets close to a

IMPORTANT POINT:
The limit value doesn't depend on f(a)!

Example: f(x) = (x² - 1)/(x - 1)
f(1) is undefined (0/0), BUT
lim[x→ 1] f(x) = lim[x→ 1] (x + 1) = 2

RIGHT vs LEFT LIMITS:
lim[x→a+] f(x): Approaching from right
lim[x→a-] f(x): Approaching from left

For limit to exist, left and right limits must be equal

DEFINITION:
A function is continuous at x = a if:
1. f(a) is defined
2. lim[x→a] f(x) exists
3. lim[x→a] f(x) = f(a)

GEOMETRIC MEANING:
You can draw the function without lifting your pencil

COMMON CONTINUITIES:✓
- Polynomials: Continuous everywhere
- Sine, cosine: Continuous everywhere
- Rational functions: Continuous except where denominator = 0
- Root functions: Continuous in their domain

DISCONTINUITIES:
- Removable: Can be fixed by redefining point
- Jump: Left and right limits differ
- Infinite: Vertical asymptote

INTUITIVE DEFINITION:
The derivative measures how fast a function changes

MATHEMATICAL DEFINITION:
f'(x) = lim[h→ 0] [f(x+h) - f(x)] / h

GEOMETRIC MEANING:
Slope of tangent line to curve at point x

PHYSICAL MEANING:
- If f(t) is position, f'(t) is velocity
- If f(t) is velocity, f'(t) is acceleration

NOTATIONS FOR DERIVATIVE:
- f'(x)
- df/dx
- [dy/dx] (Leibniz notation)

DERIVATIVES OF COMMON FUNCTIONS:

Polynomials:
- d/dx[xⁿ] = n×x^(n-1)
- d/dx[5x³] = 15x²

Trigonometric:
- d/dx[sin(x)] = cos(x)
- d/dx[cos(x)] = -sin(x)
- d/dx[tan(x)] = sec²(x)

Exponential & Logarithmic:
- d/dx[e^x] = e^x
- d/dx[ln(x)] = 1/x
- d/dx[a^x] = a^x × ln(a)

PRODUCT RULE:
If f(x) = u(x) × v(x), then
f'(x) = u'v + uv'

Exampl: f(x) = x² × sin(x)
f'(x) = 2x × sin(x) + x² × cos(x)

QUOTIENT RULE:
If f(x) = u(x)/v(x), then
f'(x) = [u'v - uv'] / v²

Example: f(x) = (x² + 1)/(x - 1)
f'(x) = [2x(x-1) - (x²+1)] / (x-1)²

CHAIN RULE:
If f(x) = g(h(x)), then
f'(x) = g'(h(x)) × h'(x)

Example: f(x) = (5x + 2)³
Let u = 5x + 2, then f = u³
f'(x) = 3u² × 5 = 3(5x+2)² × 5

CRITICAL POINTS:
Points where f'(x) = 0 or f'(x) undefined
These are candidates for maxima/minima

FIRST DERIVATIVE TEST:
At critical point x = c:
- f'(x) changes from positive to negative → MAXIMUM
- f'(x) changes from negative to positive → MINIMUM
- f'(x) doesn't change sign → INFLECTION POINT

SECOND DERIVATIVE TEST:
At critical point x = c:
- f''(c) < 0 → MAXIMUM
- f''(c) > 0 → MINIMUM
- f''(c) = 0 → Inconclusive (use 1st derivative test)

PAGE ANALYSIS:
- f'(x) > 0: Function increasing
- f'(x) < 0: Function decreasing
- f''(x) > 0: Concave up (like ∬)
- f''(x) < 0: Concave down (like ∭)

APPLICATIONS:
- Profit maximization
- Cost minimization
- Optimal dimensions
- Maximum range and height (projectile motion)

STEPS TO FIND MAX/MIN:
1. Find f'(x)
2. Set f'(x) = 0, solve for x (critical points)
3. Find f''(x) or use 1st derivative test
4. Evaluate f(x) at critical points and endpoints
5. Identify maximum and minimum values

DEFINITION:
If F'(x) = f(x), then F(x) is antiderivative of f(x)

INDEFINITE INTEGRAL:
∫ f(x)dx = F(x) + C

Where C is the constant of integration

BASIC INTEGRALS:
∫ xⁿ dx = [x^(n+1)]/(n+1) + C (n ≠ -1)
∫ 1/x dx = ln|x| + C
∫ e^x dx = e^x + C
∫ sin(x)dx = -cos(x) + C
∫ cos(x)dx = sin(x) + C

INTEGRATION RULES:

Sum/Difference:
∫ [f(x) + g(x)]dx = ∫ f(x)dx + ∫ g(x)dx

Constant Multiple:
∫ k×f(x)dx = k∫ f(x)dx

Power Rule:
∫ xⁿ dx = [x^(n+1)]/(n+1) + C

DEFINITION:
∫[a to b] f(x)dx = F(b) - F(a)

Where F is any antiderivative of f

GEOMETRIC MEANING:
Area between curve and x-axis from x = a to x = b
Note: Negative area if curve is below x-axis

PROPERTIES:
∫[a to a] f(x)dx = 0
∫[a to b] f(x)dx = -∫[b to a] f(x)dx
∫[a to c] f(x)dx + ∫[c to b] f(x)dx = ∫[a to b] f(x)dx

EXAMPLE CALCULATION:
∫[0 to 2] (3x² + 2x) dx

F(x) = x³ + x² (antiderivative)
F(2) = 8 + 4 = 12
F(0) = 0 + 0 = 0
∫[0 to 2] = 12 - 0 = 12

Area under curve from 0 to 2 is 12 square units

SUBSTITUTION METHOD (u-substitution):
Used for composite functions

Steps:
1. Choose u = some part of integrand
2. Find du
3. Rewrite integral in terms of u
4. Integrate
5. Substitute back

Example: ∫ 2x(x² + 1)³ dx
Let u = x² + 1, then du = 2x dx
∫ u³ du = u⁴/4 + C = (x²+1)⁴/4 + C

PARTS METHOD (Integration by Parts):
For products of different function types
∫ u dv = uv - ∫ v du

Mnemonic: "LIATE" (order to choose u):
Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential

Example: ∫ x×sin(x) dx
Let u = x, dv = sin(x)dx
du = dx, v = -cos(x)
∫ = -x×cos(x) - ∫ -cos(x)dx
   = -x×cos(x) + sin(x) + C

AREA BETWEEN CURVES:
If f(x) > g(x) from a to b, then
Area = ∫[a to b] [f(x) - g(x)]dx

VOLUME OF REVOLUTION:
Disc method (rotating around x-axis):
V = π ∫[a to b] [f(x)]² dx

VELOCITY FROM ACCELERATION:
If a(t) is acceleration, then
v(t) = ∫ a(t)dt

DISPLACEMENT FROM VELOCITY:
If v(t) is velocity, then
s(t) = ∫ v(t)dt

Problem: A box with open top is made from 12 m² of material.
Find dimensions that maximize volume.

If base is square (side x), height h:
4xh + x² = 12 (surface area)
h = (12 - x²)/(4x)
V = x²h = x²(12 - x²)/(4x) = (12x - x³)/4

To maximize: dV/dx = 0
(12 - 3x²)/4 = 0
12 = 3x²
x = 2 m
Then h = 1 m
Maximum volume = 4 m³

Problem: Find area between y = x² and y = 4x - 3

First find intersection points:
x² = 4x - 3
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1 or x = 3

4x - 3 > x² from 1 to 3

Area = ∫[1 to 3] [(4x-3) - x²]dx
     = [2x² - 3x - x³/3]₁³
     = (18 - 9 - 9) - (2 - 3 - 1/3)
     = 0 - (-4/3) = 4/3 square units

DERIVATIVES:
d/dx[xⁿ] = nx^(n-1)
d/dx[e^x] = e^x
d/dx[ln(x]] = 1/x
d/dx[sin(x)] = cos(x)

RULES:
Product: (uv)' = u'v + uv'
Quotient: (u/v)' = (u'v - uv')/v²
Chain: [g(h(x))]' = g'(h)×h'

INTEGRATION:
∫ xⁿ dx = x^(n+1)/(n+1) + C
∫ 1/x dx = ln|x| + C
∫ e^x dx = e^x + C
∫ sin(x)dx = -cos(x) + C