Rolle's Theorem is a foundational result that gives conditions under which a differentiable function must have a horizontal tangent line.
Let f(x) be a function that satisfies three conditions:Then, there exists at least one number c in the open interval (a, b) such that f'(c) = 0.
- f(x) is continuous on the closed interval [a, b].
- f(x) is differentiable on the open interval (a, b).
- f(a) = f(b) (the function has the same value at the endpoints).
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).
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.
Let f(x) be a function that satisfies two conditions:Then, there exists at least one number c in the open interval (a, b) such that: f'(c) = [f(b) - f(a)] / [b - a]
- f(x) is continuous on the closed interval [a, b].
- f(x) is differentiable on the open interval (a, b).
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.
This theorem involves two functions and is used to prove L'Hospital's Rule.
Let f(x) and g(x) be two functions that satisfy: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)
- f(x) and g(x) are continuous on the closed interval [a, b].
- f(x) and g(x) are differentiable on the open interval (a, b).
- g'(x) ≠ 0 for all x in (a, b).
Note: If we let g(x) = x, then g'(x) = 1, and this theorem reduces to Lagrange's MVT.
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.
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)
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.
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.
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! + ...
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! + ...
You must memorize the following standard Maclaurin series expansions.
| Function | Maclaurin's Series Expansion | Interval of Convergence |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + x⁴/4! + ... | 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 + ... | -1 < x ≤ 1 |
| (1 + x)ᵐ | 1 + mx + [m(m-1)/2!]x² + [m(m-1)(m-2)/3!]x³ + ... | |x| < 1 |
| aˣ | ex*log(a) = 1 + x(log a) + [x²(log a)²]/2! + ... | For all x |
This involves using derivatives to find the "highest" (maximum) and "lowest" (minimum) points on a function's graph.
Critical Points: Points 'c' where f'(c) = 0 or f'(c) is undefined. These are candidates for maxima or minima.
This test finds local maxima and minima by checking the sign of f'(x) around a critical point c.
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.
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.
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 ±∞).
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.
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).
Problem: lim (x → a) [f(x)]ᵍ(ˣ)
Solution: