Knowlet

Unit 3: Mean Value Theorems and Expansions

Rolle's Theorem

Rolle's Theorem is a special case of the Mean Value Theorem. It gives conditions under which a differentiable function must have a horizontal tangent line.

Statement

Let f(x) be a function that satisfies three conditions:
  1. f(x) is continuous on the closed interval [a, b].
  2. f(x) is differentiable on the open interval (a, b).
  3. f(a) = f(b) (the function has the same value at the endpoints).
Then, there exists at least one number c in the open interval (a, b) such that f'(c) = 0.

Geometrical Interpretation

If a continuous and smooth curve starts and ends at the same height (f(a) = f(b)), there must be at least one point 'c' between 'a' and 'b' where the tangent line is horizontal (slope = 0).

Exam Tip: To "verify" Rolle's Theorem for a function f(x) on [a, b], you must:
  1. Check continuity on [a, b]. (Polynomials, sin, cos, eˣ are always continuous).
  2. Check differentiability on (a, b). (Find f'(x) and see if it's defined).
  3. Check if f(a) = f(b).
  4. If all three hold, solve the equation f'(x) = 0 to find 'c'.
  5. Show that this 'c' value lies strictly between 'a' and 'b' (i.e., a < c < b).

Lagrange's Mean Value Theorem (MVT)

This is a more general version of Rolle's Theorem. It states that the slope of the secant line between two points is equal to the slope of the tangent line at some intermediate point.

Statement

Let f(x) be a function that satisfies two conditions:
  1. f(x) is continuous on the closed interval [a, b].
  2. f(x) is differentiable on the open interval (a, b).
Then, there exists at least one number c in the open interval (a, b) such that: f'(c) = [f(b) - f(a)] / [b - a]

Geometrical Interpretation

The term [f(b) - f(a)] / [b - a] is the slope of the secant line connecting the endpoints (a, f(a)) and (b, f(b)). The term f'(c) is the slope of the tangent line at x = c.

The theorem guarantees that there is at least one point 'c' where the tangent line is parallel to the secant line connecting the endpoints.

Cauchy's Mean Value Theorem (Generalized MVT)

This theorem involves two functions and is used to prove L'Hospital's Rule.

Statement

Let f(x) and g(x) be two functions that satisfy:
  1. f(x) and g(x) are continuous on the closed interval [a, b].
  2. f(x) and g(x) are differentiable on the open interval (a, b).
  3. g'(x) ≠ 0 for all x in (a, b).
Then, there exists at least one number c in the open interval (a, b) such that: [f(b) - f(a)] / [g(b) - g(a)] = f'(c) / g'(c)

Note: If we let g(x) = x, then g'(x) = 1, g(b) = b, and g(a) = a. The formula becomes [f(b) - f(a)] / [b - a] = f'(c) / 1, which is exactly Lagrange's MVT. Thus, Lagrange's MVT is a special case of Cauchy's MVT.

Generalized MVT (Taylor's Series in Finite Form)

This is a more powerful generalization, also known as Taylor's Theorem with Remainder. It allows us to approximate a function f(x) near a point 'a' using a polynomial, and it gives a formula for the error (remainder) in that approximation.

Taylor's Theorem with Lagrange's Form of Remainder

If a function f(x) and its first n derivatives (f', f'', ..., fⁿ) are continuous on [a, b] and fⁿ(x) is differentiable on (a, b), then for any x in [a, b]: f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)²/2! + ... + fⁿ⁻¹(a)(x-a)ⁿ⁻¹/(n-1)! + Rₙ Where Rₙ is the Lagrange Form of the Remainder, given by: Rₙ = fⁿ(c)(x-a)ⁿ / n! for some number c between 'a' and 'x'.

Taylor's Theorem with Cauchy's Form of Remainder

Under the same conditions, the Cauchy Form of the Remainder is: Rₙ = fⁿ(c)(x-c)ⁿ⁻¹(x-a) / (n-1)! for some number c between 'a' and 'x'.

Lagrange's MVT is just Taylor's Theorem for n = 1.

Expansion of Functions in Infinite Power Series

If the remainder term Rₙ from Taylor's Theorem approaches 0 as n approaches infinity (lim (n → ∞) Rₙ = 0), then the function can be represented exactly by an infinite power series.

Taylor's Series (about x = a)

The Taylor series expansion of f(x) about the point x = a is: f(x) = Σ [fⁿ(a) / n!] * (x-a)ⁿ (from n=0 to ∞) f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ...

Maclaurin's Series (about x = 0)

This is just a special case of the Taylor series where a = 0.

The Maclaurin series expansion of f(x) is: f(x) = Σ [fⁿ(0) / n!] * xⁿ (from n=0 to ∞) f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ...
Exam Tip: You MUST memorize the standard Maclaurin series for:
  • eˣ = 1 + x + x²/2! + x³/3! + ... (for all x)
  • sin x = x - x³/3! + x⁵/5! - x⁷/7! + ... (for all x)
  • cos x = 1 - x²/2! + x⁴/4! - x⁶/6! + ... (for all x)
  • ln(1 + x) = x - x²/2 + x³/3 - x⁴/4 + ... (for -1 < x ≤ 1)
  • (1 + x)ᵐ = 1 + mx + m(m-1)x²/2! + ... (Binomial series, for |x| < 1)
These are often used to evaluate limits without L'Hospital's Rule.

Example: Evaluate lim (x → 0) (eˣ - 1 - x) / x²

  1. Use the Maclaurin series for eˣ: eˣ = 1 + x + x²/2! + x³/3! + ...
  2. Substitute into the limit: lim (x → 0) [ (1 + x + x²/2! + x³/3! + ...) - 1 - x ] / x²
  3. Simplify the numerator: lim (x → 0) [ (x²/2! + x³/3! + ...) ] / x²
  4. Factor out x² from the numerator and cancel: lim (x → 0) [ x²(1/2! + x/3! + ...) ] / x² lim (x → 0) [ 1/2! + x/3! + x²/4! + ... ]
  5. Now, substitute x = 0: 1/2! + 0 + 0 + ... = 1/2

Increasing and Decreasing Functions

This is a re-statement of the topic from Unit 2, now formally provable using the Mean Value Theorem.

Let f(x) be differentiable on (a, b).

  • If f'(x) > 0 for all x in (a, b), then f(x) is strictly increasing on (a, b).
  • If f'(x) < 0 for all x in (a, b), then f(x) is strictly decreasing on (a, b).

Critical Points: Points 'c' where f'(c) = 0 or f'(c) is undefined. These are the only points where the function can change from increasing to decreasing (or vice versa).

First Derivative Test

This test finds local maxima and minima by checking the sign of f'(x) around a critical point c.

  • If f'(x) changes from positive (+) to negative (-) at c, then f(x) has a local maximum at c.
  • If f'(x) changes from negative (-) to positive (+) at c, then f(x) has a local minimum at c.
  • If f'(x) does not change sign at c, then c is a point of inflection (e.g., f(x) = x³ at c = 0).

Maxima and Minima for Functions of Single Variables

This topic involves finding the "highest" (maximum) and "lowest" (minimum) points on a function's graph.

Second Derivative Test

This is often a faster way to classify critical points, but it only works if f''(x) is easy to find and f''(c) ≠ 0.

Let 'c' be a critical point where f'(c) = 0.

  1. Find the second derivative, f''(x).
  2. Evaluate f''(c):
    • If f''(c) < 0 (concave down), f(x) has a local maximum at x = c.
    • If f''(c) > 0 (concave up), f(x) has a local minimum at x = c.
    • If f''(c) = 0, the test is inconclusive. You must go back and use the First Derivative Test.

Absolute (Global) Maxima and Minima on a Closed Interval [a, b]

To find the absolute highest and lowest values of a continuous function f(x) on [a, b]:

  1. Find all critical points of f(x) that are inside the interval (a, b).
  2. Make a list of "candidate" points:
    • The critical points from step 1.
    • The endpoints of the interval, a and b.
  3. Evaluate the function f(x) at every point in your candidate list.
  4. The largest value is the absolute maximum.
  5. The smallest value is the absolute minimum.
Common Mistake: Forgetting to check the endpoints! The absolute max or min on a closed interval can often occur at the endpoints (e.g., f(x) = x on [1, 2]).

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