Unit 2: Mean Value Theorems, Expansions, and Extrema

Mean Value Theorems

Mean Value Theorems are fundamental results in calculus that connect the local behavior of a function (its derivative) to its global behavior (its values at the endpoints of an interval).

Rolle's Theorem

This is the simplest of the mean value theorems and serves as a basis for the others.

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 values at the endpoints are equal).

Then, there exists at least one point c in the open interval (a, b) such that f'(c) = 0.

Geometrical Meaning: If a smooth, continuous curve starts and ends at the same height, there must be at least one point between them where the tangent line is horizontal.

Lagrange's Mean Value Theorem (MVT)

This is a generalization of Rolle's Theorem. It removes the condition that f(a) = f(b).

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 point c in the open interval (a, b) such that:
f'(c) = [f(b) - f(a)] / [b - a]

Geometrical Meaning: The term on the right, [f(b) - f(a)] / [b - a], is the slope of the **secant line** connecting the endpoints (a, f(a)) and (b, f(b)). The theorem states that there must be at least one point c where the **tangent line** is parallel to this secant line.

Cauchy's Mean Value Theorem (CMVT)

This is a further generalization, also known as the "Extended MVT". It deals with two functions simultaneously and is used to prove L'Hôpital's Rule.

Statement: Let f(x) and g(x) be two functions that satisfy:

1. f(x) and g(x) are continuous on [a, b].
2. f(x) and g(x) are differentiable on (a, b).
3. g'(x) ≠ 0 for any x in (a, b).

Then, there exists at least one point c in (a, b) such that:
[f'(c)] / [g'(c)] = [f(b) - f(a)] / [g(b) - g(a)]

Note: If you let g(x) = x, then g'(x) = 1 and g(b) - g(a) = b - a. This reduces CMVT directly to Lagrange's MVT.

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Taylor's and Maclaurin's Theorems

These theorems provide a way to approximate a complex, differentiable function near a point using a simpler polynomial.

Taylor's Theorem (Statement)

Statement: If a function f(x) is n-times differentiable at a point a, it can be approximated by a polynomial of degree n.

The **Taylor series expansion** of f(x) about the point x = a is:

f(x) = f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)² + [f'''(a)/3!](x-a)³ + ... + [f⁽ⁿ⁾(a)/n!](x-a)ⁿ + ...

This is an infinite series. The syllabus also mentions "Taylor's theorem," which often refers to the finite form with a remainder term, but the series expansion is the primary application.

Maclaurin's Theorem and Series

A **Maclaurin series** is just a special case of the Taylor series where the expansion is centered at the point a = 0.

f(x) = f(0) + f'(0)x + [f''(0)/2!]x² + [f'''(0)/3!]x³ + ... + [f⁽ⁿ⁾(0)/n!]xⁿ + ...

Standard Expansions

These are derived by applying the Maclaurin series formula and are essential to memorize.

• 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)
• log(1 + x) = x - x²/2 + x³/3 - x⁴/4 + ... (for -1 < x ≤ 1)

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Maxima and Minima

This is the process of finding the local "peaks" (maxima) and "valleys" (minima) of a function f(x) using derivatives.

Necessary and Sufficient Conditions

Step 1: Find Critical Points (Necessary Condition)

A critical point is a point x = c where the function *could* have a max or min. This occurs where the tangent is horizontal.

• Find the first derivative: f'(x).
• Set it to zero and solve for x: f'(x) = 0.

The solutions x = c₁, c₂, ... are the critical points.

Step 2: Test the Critical Points (Sufficient Conditions)

There are two tests to determine if a critical point c is a maximum, minimum, or neither.

A) The First Derivative Test

1. Check the sign of f'(x) just *before* c (at c - h) and just *after* c (at c + h).
2. If the sign changes from (+) to (-): The function was increasing, then decreasing. This is a Local Maximum.
3. If the sign changes from (-) to (+): The function was decreasing, then increasing. This is a Local Minimum.
4. If there is no sign change: It is a point of inflection (e.g., f(x) = x³ at x = 0).

B) The Second Derivative Test (Often easier)

1. Find the second derivative: f''(x).
2. Evaluate f''(x) *at* the critical point c.
3. If f''(c) < 0 (Negative): The function is "concave down" (like a frown). This is a Local Maximum.
4. If f''(c) > 0 (Positive): The function is "concave up" (like a smile). This is a Local Minimum.
5. If f''(c) = 0: The test is inconclusive. You must go back and use the First Derivative Test.

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Indeterminate Forms

These are limits that cannot be determined by simple substitution, as they result in ambiguous forms like 0/0 or ∞/∞.

L'Hôpital's Rule (0/0 and ∞/∞ forms)

Rule: 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 second limit exists.

Example (0/0): lim (x→0) sin(x) / x
= lim (x→0) cos(x) / 1 (by L'Hôpital's)
= cos(0) / 1 = 1

Other Forms (0×∞, ∞ - ∞, 0⁰, 1⁰⁰, ∞⁰)

The key is to algebraically convert these forms into 0/0 or ∞/∞ so you can use L'Hôpital's Rule.

Form 0 × ∞: lim f(x)g(x) where f→0, g→∞

• Method: Rewrite f·g as f / (1/g) (now 0/0) or g / (1/f) (now ∞/∞).
• Example: lim (x→0⁺) x log(x) (0 × -∞)
  = lim (x→0⁺) log(x) / (1/x) (now -∞/∞)
  = lim (x→0⁺) (1/x) / (-1/x²) (by L'Hôpital's)
  = lim (x→0⁺) -x = 0

Form ∞ - ∞:

• Method: Find a common denominator or use rationalization to combine the terms into a single fraction.
• Example: lim (x→0) (1/x - 1/sin(x)) (∞ - ∞)
  = lim (x→0) (sin(x) - x) / (x sin(x)) (now 0/0)
  ...then apply L'Hôpital's Rule (multiple times).

Exponential Forms (0⁰, 1⁰⁰, ∞⁰):

• Method: Let L be the limit. Take the natural logarithm (ln) of both sides.
• Let y = f(x)ᵍ⁽ˣ⁾. Then ln(y) = g(x) ln(f(x)).
• This turns the limit into a 0 × ∞ form.
• Example: L = lim (x→0⁺) xˣ (0⁰ form)
  Let y = xˣ, so ln(y) = x log(x)
  We need to find lim (x→0⁺) ln(y) = lim (x→0⁺) x log(x)
  From the example above, we know this limit is 0.
  So, ln(L) = 0.
  To find L, we exponentiate: L = e⁰ = 1.

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Unit 2: Exam Quick Tips