Unit 2: Mean Value Theorems, Expansions, and Indeterminate Forms

Rolle's Theorem

Rolle's Theorem is a foundational result that 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).

Lagrange's Mean Value Theorem (MVT)

This is a generalization 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, and this theorem reduces to Lagrange's MVT.

Taylor's and Maclaurin's Theorems (Statement and Applications)

These theorems, stated without proof, provide a way to approximate a function f(x) near a point 'a' using a polynomial. The formula includes a "remainder" term (Rₙ) that quantifies the error of the approximation.

Taylor's Theorem (with Lagrange's Form of Remainder)

If a function f(x) is such that its first (n-1) derivatives are continuous on [a, a+h] and its n-th derivative fⁿ(x) exists on (a, a+h), then there exists at least one number θ (where 0 < θ < 1) such that: f(a+h) = f(a) + h*f'(a) + (h²/2!)*f''(a) + ... + (hⁿ⁻¹/(n-1)!)*fⁿ⁻¹(a) + Rₙ Where Rₙ (the remainder) is given by: Rₙ = (hⁿ/n!) * fⁿ(a + θh)

Maclaurin's Theorem

This is just a special case of Taylor's Theorem where a = 0 and h = x.

f(x) = f(0) + x*f'(0) + (x²/2!)*f''(0) + ... + (xⁿ⁻¹/(n-1)!)*fⁿ⁻¹(0) + Rₙ
Where Rₙ (the remainder) is given by: Rₙ = (xⁿ/n!) * fⁿ(θx), for 0 < θ < 1.

Taylor's and Maclaurin's Series

If the remainder term Rₙ from Taylor's or Maclaurin'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 the most common form. It is the Taylor series with 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! + ...

Expansion of Standard Functions

You must memorize the following standard Maclaurin series expansions.

Maxima and Minima for Functions of One Variable

This involves using derivatives to find the "highest" (maximum) and "lowest" (minimum) points on a function's graph.

Necessary and Sufficient Conditions

Critical Points: Points 'c' where f'(c) = 0 or f'(c) is undefined. These are candidates for maxima or minima.

First Derivative Test (Necessary Condition)

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, c is a point of inflection.

Second Derivative Test (Sufficient Condition)

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 use the First Derivative Test (or higher-order derivatives).

Indeterminate Forms

When evaluating a limit, we sometimes get an undefined expression. These are called indeterminate forms. L'Hospital's Rule is the primary tool to solve them.

L'Hospital's Rule (for 0/0 and ∞/∞)

If lim (x → a) f(x) / g(x) results in the form 0/0 or ∞/∞,
Then: lim (x → a) [f(x) / g(x)] = lim (x → a) [f'(x) / g'(x)] ...provided the limit on the right side exists (or is ±∞).

Other Indeterminate Forms (and how to convert them)

• Form 0 · ∞:

  Problem: lim (x → a) f(x) * g(x) (where f→0, g→∞)
  Solution: Rewrite as f(x) / (1/g(x)) (now 0/0) or g(x) / (1/f(x)) (now ∞/∞).
  Example: lim (x → 0⁺) x * log(x)
  Rewrite as lim (x → 0⁺) log(x) / (1/x). This is now -∞/∞. Apply L'Hospital's Rule.
  = lim (x → 0⁺) [ (1/x) / (-1/x²) ] = lim (x → 0⁺) -x = 0.

• Form ∞ - ∞:

  Problem: lim (x → a) [f(x) - g(x)]
  Solution: Use algebra to combine the terms into a single fraction (e.g., find a common denominator, or rationalize). This will usually turn it into 0/0 or ∞/∞.
  Example: lim (x → 0⁺) [ (1/x) - (1/sin x) ]
  Combine: lim (x → 0⁺) [ (sin x - x) / (x * sin x) ]. This is now 0/0. Apply L'Hospital's Rule (possibly multiple times).

• Exponential Forms (0⁰, 1^∞, ∞⁰):

  Problem: lim (x → a) [f(x)]ᵍ(ˣ)
  Solution:

  1. Let L be the limit: L = lim (x → a) [f(x)]ᵍ(ˣ)
  2. Take the natural logarithm (ln) of both sides: ln L = ln [ lim (x → a) [f(x)]ᵍ(ˣ) ] ln L = lim (x → a) [ ln( [f(x)]ᵍ(ˣ) ) ] ln L = lim (x → a) [ g(x) * ln(f(x)) ]
  3. This new limit is now a 0 · ∞ form. Convert it to 0/0 or ∞/∞ as shown above and solve it. Let the result be K.
  4. Final Answer: You have found ln L = K. The original limit is L = eᴷ.