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).
This is the simplest of the mean value theorems and serves as a basis for the others.
Statement: Letf(x)be a function that satisfies three conditions:Then, there exists at least one point
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 values at the endpoints are equal).cin the open interval(a, b)such thatf'(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.
This is a generalization of Rolle's Theorem. It removes the condition that f(a) = f(b).
Statement: Letf(x)be a function that satisfies two conditions:Then, there exists at least one point
f(x)is continuous on the closed interval[a, b].f(x)is differentiable on the open interval(a, b).cin 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.
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: Letf(x)andg(x)be two functions that satisfy:Then, there exists at least one point
f(x)andg(x)are continuous on[a, b].f(x)andg(x)are differentiable on(a, b).g'(x) ≠ 0for anyxin(a, b).cin(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.
These theorems provide a way to approximate a complex, differentiable function near a point using a simpler polynomial.
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.
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 + ...
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)This is the process of finding the local "peaks" (maxima) and "valleys" (minima) of a function f(x) using derivatives.
A critical point is a point x = c where the function *could* have a max or min. This occurs where the tangent is horizontal.
f'(x).f'(x) = 0.The solutions x = c₁, c₂, ... are the critical points.
There are two tests to determine if a critical point c is a maximum, minimum, or neither.
A) The First Derivative Test
f'(x) just *before* c (at c - h) and just *after* c (at c + h).f(x) = x³ at x = 0).B) The Second Derivative Test (Often easier)
f''(x).f''(x) *at* the critical point c.f''(c) < 0 (Negative): The function is "concave down" (like a frown). This is a Local Maximum.f''(c) > 0 (Positive): The function is "concave up" (like a smile). This is a Local Minimum.f''(c) = 0: The test is inconclusive. You must go back and use the First Derivative Test.These are limits that cannot be determined by simple substitution, as they result in ambiguous forms like 0/0 or ∞/∞.
Rule: Iflim (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
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→∞
f·g as f / (1/g) (now 0/0) or g / (1/f) (now ∞/∞).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 = 0Form ∞ - ∞:
lim (x→0) (1/x - 1/sin(x)) (∞ - ∞)
lim (x→0) (sin(x) - x) / (x sin(x)) (now 0/0)
Exponential Forms (0⁰, 1⁰⁰, ∞⁰):
L be the limit. Take the natural logarithm (ln) of both sides.y = f(x)ᵍ⁽ˣ⁾. Then ln(y) = g(x) ln(f(x)).L = lim (x→0⁺) xˣ (0⁰ form)
y = xˣ, so ln(y) = x log(x)
lim (x→0⁺) ln(y) = lim (x→0⁺) x log(x)
ln(L) = 0.
L = e⁰ = 1.y = f(x)ᵍ⁽ˣ⁾, 2. Take ln, 3. Find lim ln(y) (which will be 0×∞ form), 4. Solve that limit (call it L*), 5. The final answer is e^(L*). Do not forget the last step!f'' < 0 (frown) is Max. f'' > 0 (smile) is Min. If f'' = 0, the test fails, and you must use the 1st Derivative Test.eˣ, sin(x), cos(x), and log(1+x).