Functions, Limits, Derivatives

Lecture Outline

• This lecture covers functions, analytical solutions to polynomial equations, limits, function continuity, derivatives, monotonicity, concavity and practical examples.

Functions

• A function relates or maps values from one set (domain) to another (range).
• The domain is the input set, and the range is the output set of a function.
• Functions can be univariate (one input variable) or multivariate (multiple input variables).
• Univariate, real-valued functions are notated f : D ⊆ R → R indicating that for all x in D, f(x) is in R.
• The root of a function f is the value x in its domain D, such that f(x) = 0.

Function Evaluations

• Functions can be evaluated by substituting values into the equation.
• For the function f(x) = 11x + 7:
  • f(0) = 11 * 0 + 7 = 7.
  • f(3) = 11 * 3 + 7 = 40.
  • f(-5) = 11 * -5 + 7 = -48.
• For the function g(x) = x² - 4:
  • g(0) = (0)² - 4 = -4.
  • g(2) = (2)² - 4 = 0.
  • g(-2) = (-2)² - 4 = 0.

Compositions

• Given f(x) = 2x² + 3x + 5 and g(x) = x + 1:
  • The compositon f(g(x)) = 2(x + 1)² + 3(x + 1) + 5 = 2x² + 7x + 10.
  • The composition g(g(x)) = (x + 1) + 1 = x + 2.
  • The composition f(f(x)) = 2(2x² + 3x + 5)² + 3(2x² + 3x + 5) + 5 = 8x⁴ + 28x³ + 58x² + 69x + 70.
  • The composition g(f(x)) = (2x² + 3x + 5) + 1 = 2x² + 3x + 6

Transcendental Functions

• Trigonometric Functions: sin(x), cos(2x - 5), tan(3x - 2), cot(x - 3), arctan(x), arccos(x + 2).
• Exponential Functions: ex, e2x+4
• Logarithmic Functions: ln(x), ln(sin(x) + cos(x)).

Piecewise Functions

• Piecewise functions are defined by different expressions over different intervals for f(x):
  • x² - 1 when x ≥ 0
  • sin(x) when x > 0
• Piecewise functions are defined by different expressions over different intervals for g(x):
  • -3x when x > 1
  • -2 when x = 1
  • x/2 when x < 1

Analytical Equation Solving - Linear Equations

• Linear equations can be solved to find x.
• For example, solving 2x + 7 = 7x + 2 involves isolating x to find that x = -1.
• Solving 5(x - 1) = 3(2x - 5) - (1 - 3x) leads to x = 9/4.
• Given the function f(x) = 2(x - 1) - 6x:
  • To find x, where f(x) = 6, solve 2(x - 1) - 6x = 6.
  • This simplifies to x = -2.

Analytical Equation Solving - Quadratic Equations

• Quadratic equations (ax² + bx + c = 0) can be solved using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)
• For the equation x² + 5x + 6 = 0:
  • The quadratic formula is applied with a = 1, b = 5, and c = 6.
  • Solutions are x = -2 and x = -3.
• Application Example: Given revenue function f(x) = 1000 - 5x² - 20x and an initial investment of 970:
  • To find when revenue equals the initial investment, set f(x) = 970.
  • The company will get the required revenue in 6 weeks.

Analytical Equation Solving - Cubic Equations

• Cubic equations such as x³ - 6x² + 11x = 6 can be solved by emplying factorization rules.
• Factorization can simplify the equation down to linear terms.
• Solve x³ - 6x² + 11x - 6 = 0:
  • Use identity a³ – b³ = (a - b)(a² + ab + b²) to solve
  • The solutions are x = 1, x = 3, and x = 2.

Analytical Equation Solving - Generic Equations

• Equations like 5x⁷ + 4x⁵ + 2x³ + 3x² + 4 = 0, ex + cosx = 0, 2x + x³ + 5 = 0, or lnx + x² + 7 = 0 are difficult to solve analytically.
• Hybrid transcendental equations typically lack analytical solutions.
• Module focus is using numerical methods for these equations.

Limits

• The limit of f(x) as x approaches a number a is L, indicating f(x) gets arbitrarily close to L as x gets sufficiently close to a, without necessarily equalling a
• The notation for a limit is expressed as limx→α f(x) = L.

Limits Evaluation

• Evaluate f(x) = 1/(x - 1) by approaching a specific point from both sides.
• Investigate the behavior of f(x) as x approaches a specific point from both sides.
• If the estimated value from both sides is the same, the limit exists, but note x ≠ α in all cases.
  • Right-side limit: limx→α+ f(x) = L
  • Left-side limit: limx→α- f(x) = L
• Limits can be infinite.
• It is of interest to see what happens when x approaches infinity, for example, limx→+∞ x = 0.

Limit Properties

• The limit of a constant times a function is the constant times the limit of the function: limx→α cf(x) = c * limx→α f(x)
• The limit of a sum is the sum of the limits: limx→α [f(x) + g(x)] = limx→α f(x) + limx→α g(x)
• The limit of a product is the product of the limits: limx→α [f(x) * g(x)] =limx→α f(x) * limx→α g(x)
• The limit of a quotient is the quotient of the limits, when the limit of the denominator is nonzero: limx→α f(x)/g(x) = limx→α f(x) / limx→α g(x)
• The limit of a function raised to a power is the limit of the function, raised to the power: limx→α [f(x)]^n = [limx→α f(x)]^n
• The limit of the nth root of a function is the nth root of the limit of the function: limx→α n√f(x) = ⁿ√limx→α f(x)
• The limit of a constant is the constant: limx→α c = c
• The limit of x as x approaches α is α: limx→α x = α
• The limit of x^n as x approaches α is α^n: limx→α x^n = α^n
• If f(x) ≤ g(x) for all x in [α, β] (except possibly at x = c), then limx→c f(x) ≤ limx→c g(x).

Continuity

• A function f(x) is continuous at x = α if the limit of f(x) as x approaches α from both sides exists, and equals f(α).
• Expressed as: limx→α- f(x) = limx→α+ f(x) = limx→α f(x) = f(α)
• A function is continuous on an interval [α, β] if it is continuous at each point within that interval.

Intermediate Value Theorem

• If a function f(x) is continuous on [α, β], and M is a value between f(α) and f(β), then there exists a c between α and β such that f(c) = M.
• The Intermediate Value Theorem is also known as Bolzano's Theorem.
• The theorem is valuable for proving the roots over an interval.

Derivative

• The derivative of a function f(x) , f'(x) represents the rate of change with respect to x, and it is defined as: f'(x) = limh→0 [f(x + h) - f(x)] / h
• A function is differentiable at a point x=α is its derivative, f'(α), exists.
• The velocity of a object is defined as the change of rate of the objects postition with respect to time.
• In single vairable functions, the derivative calculates the slope of the tangent line on a given point.

Derivative Formulas

• Summation/Subtraction Rule: (f(x) ± g(x))' = f'(x) ± g'(x)
• Product Rule: (f(x)g(x))' = f'(x)g(x) + g'(x)f(x)
• Quotient Rule: (f(x)/g(x))' = [f'(x)g(x) - g'(x)f(x)] / [g(x)]²
• Chain Rule: (f(g(x)))' = f'(g(x))g'(x)
• Scalar Multiplication: (cf(x))' = cf'(x)
• Second Derivative: (f(x))" = (f'(x))' = f"(x)

Derivatives of Common Functions

• Constant Functions: (c)' = 0
• Polynomial Functions: (cxⁿ)' = cnxⁿ⁻¹
• Exponential Functions: (eˣ)' = eˣ
• Logarithmic Functions: (lnx)' = 1/x
• Trigonometric Functions: (sinx)' = cosx, (cosx)' = -sinx

Relative Extrema

• A relative minimum of f(x) in (α, β) at c means α < c < β and f(x) ≥ f(c) for all x in (α, β).
• A relative maximum of f(x) in (α, β) at c means α < c < β and f(x) ≤ f(c) for all x in (α, β).

Critical Points

• The function f(x) has a critical point and its deritivate, f'(x) exists if:
  • f'(c) = 0
  • The point x = c and there is no f'(c)
• Fermat's Theorem:
  • If f(x) has a relative extrema at a point x= c and f'(x) exists, then x = c is a critical point of f(x), so that, f'(c) = 0.

Critical Points Example

• Locate the critical points of the function f(x) = 14x³ + 18x² - 6x - 8
  • First calculate f'(x) = 42x² + 36x - 6
  • Solve the f'(x) = 0
  • Finding the two critical points which are: x = -84/54 and x = 12/54

Critical Points Example with Fractions

• Locate the critical points of the function f(x) = (x+4)/(2x²+x+8)
  • First calculate f'(x) = (2x² + x + 8 - (4x + 1)(x + 4))/(2x² + x + 8)²
  • Find out when the nominator and denominator is zero finding the critical points which are: x = -4 ± √2

Critical Points Example with Natural Log

• Locate the critical points of the function f(x) = ln(4x² - 8x + 17)
  • First calculate f'(x) = (8x - 8)/(4x² - 8x + 17)
  • Solve f'(x) = 0, critical point x = 1

Relative Extrema Example

• Locate the relative extrema of the function f(x) = x³ - 2x + 1
  • First calculate f'(x) = 3x² - 2 = 0
  • Find the x values of the critical points, x = ±√6/3
  • We can see that −√6/3 is a local maxima and √6/3 is a local minima of the function.

Absolute Extrema

• Extreme Value Theorem:
  • If a function f is continuous on a closed interval [α, β], then f has both an absolute maximum and an absolute minimum on [α, β].

Identify Absolute Extrema

• Process in identify absolute extrema in a function:
  • Make sure the function is continous on the desired interval.
  • Find the derivative and locate the critical points of f(x).
  • Evaluate the function at the critical points and end points.
  • Correspond to the largest and smallest y-values

Absolute Extrema Example

• Find the absolute extrema of the function: f(x) = 2x³ - x² + 1 over the interval [-1, 2].
  • Check if the function meet the requirements.
  • Solve for the derivative f'(x) = 6x² - 2x = 0 -The critical points are at f have been found at x = 1/3 and x = 0
  • Evaluate the function on the critical points and on the endpoints f(-1) = -2, f(0) = 1, f(1/3) ≈ 0.96, f(2) = 13
  • The absolute maximum is -1 and the absolute minimum is: 2

Mean Value Theorem (MVT)

• Given f(x) is differentiable on (α, β) and continuous on [α, β]:
  • Then there exists the point c in (α, β) so that f'(c) = (f(b) - f(a))/(b - a)
• Rolle's Theorem: Special Case
  • Given f(x) is differentiable on (α, β) and continuous on [α, β]:
  • Given that f(α) = f(β) then there exists the point c in (α, β) so that f'(c) = 0

The Mean Value Theorem – Example

• Find all the possible values of c that satisfy the Mean Value Theorem for a function f(x) = 2 - 3x - 4x² within the interval [0, 2]
  • Step 1: Make sure the function meet the MVT requirements.
  • Find the function for the desired interval f'(x) = -11
  • Find a value c given f'(c) =-11, c = 1

The Mean Value Theorem – Another Example

• Find all the possible values of c that satisfy the Mean Value Theorem for a function f(x) = 7x³ + 4x² - 3 within the interval [1, 4]
  • Step 1: Make sure the function meet the MVT requirements.
  • Find the function for the desired interval f'(x) = 167
  • Calculate values c given that f'(x) = 21x² + 8x c ≈ 1.546

Monotonicity

• A function f(x) is increasing over an interval A, if ∀x1, x2 ∈ A, with x1 < x2 and f(x1) ≤ f(x2), and can also be refered to monotonically increasing.
• A function f(x) is decreasing over an interval A, if ∀x1, x2 ∈ A, with x1 < x2 and f(x1) ≥ f(x2), and can also be refered to monotonically decreasing.

Monotonicity Remarks

• For a function increase, if f'(x) ≥ 0, ∀x ∈ A, then f(x) is increasing over the interval A.
• For a function descrease, if f'(x) ≤ 0, ∀x ∈ A, then f(x) is decreasing over the interval A.
• For a function constance, if f'(x) = 0, ∀x ∈ A, then f(x) is constant over the interval A.
• A function can also be strictly increasing/decreasing where ∀x1, x2 ∈ A, with x1 < x2

Monotonicity First Derivative Test

• If x = c is a critical point of a function f(x):
  • If f'(x) > 0 to the left of x = c and f'(x) < 0 to the right of x = c then x = c is a relative maximum.
  • If f'(x) < 0 to the left of x = c and f'(x) > 0 to the right of x = c then x = c is a relative minimum.
  • If f'(x) is constant on both sides of x = c then x = c is neither a relative maximum nor a relative minimum.
• The notes:
  • Use the methodology for identify relative extrema and study the behavior of the first derivative of a function f.
  • Explicitly check the points where the function is not continuous (if any).

Concavity-Inflection Points

• Given f(x) is a differentiable function on the open interval (a, b):
  • f is concave up on (a, b) if f' is increasing on (a, b).
  • f is concave down on (a, b) if f' is decreasing on (a, b).
• A continuous function f(x), has a point x = c defined an inflection point when the concavity changes at that point.
• For a function to be concave, f"(x) > 0, ∀x ∈ A, and f(x) is concave up over the interval A.
• For a function to be convex, f"(x) < 0, ∀x ∈ A, and f(x) is concave down over the interval A.

Concavity - Inflection Points Second Derivative Test

• If x = c is a critical point of a function f(x) and f"(x) exists and is continuous:
  • If f"(c) < 0 then c is a relative maximum.
  • If f"(c) > 0 then c is a relative minimum.
  • If f"(c) = 0 then c can be either a relative maximum or minimum or neither.
• Use the process for identifying relative extrema by studying the behavior of the second derivative of a function f.
• Explicitly check the points where the second derivative is not continuous (if any).