Mean Value Theorem
Short Notes
The Mean Value Theorem (MVT) establishes a relationship between the average rate of change and the instantaneous rate of change of a function.
Statement: If f(x) is continuous on [a, b] and differentiable on (a, b), then there exists at least one c ∈ (a, b) such that: $\(f'(c) = \frac{f(b) - f(a)}{b - a}\)$
Geometric Interpretation: There exists a point where the tangent line is parallel to the secant line connecting (a, f(a)) and (b, f(b)).
Key Theories & Formulas
1. Rolle's Theorem (Special Case of MVT)
If f(x) is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one c ∈ (a, b) such that: $\(f'(c) = 0\)$
2. Cauchy's Mean Value Theorem (Generalized MVT)
If f(x) and g(x) are both continuous on [a, b] and differentiable on (a, b), with g'(x) ≠ 0, then there exists c ∈ (a, b) such that: $\(\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\)$
3. Applications of MVT
- Proving Inequalities: Show \(f'(x)\) bounds using MVT
- Finding Root Bounds: If f(a), f(b) have opposite signs and f is continuous
- Proving Uniqueness of Roots: Using Rolle's theorem in reverse
- Taylor's Theorem: MVT is a special case of Taylor's with n=1
4. Common Results from MVT
- If f'(x) = 0 for all x ∈ (a, b), then f is constant on [a, b]
- If f'(x) > 0 for all x ∈ (a, b), then f is strictly increasing
- If f'(x) < 0 for all x ∈ (a, b), then f is strictly decreasing
- If |f'(x)| ≤ M for all x, then |f(b) - f(a)| ≤ M|b - a|
5. Lagrange's Mean Value Theorem
Another name for MVT. The value c is called the mean value point or intermediate point.
Example Problems
Problem 1: Verify MVT for f(x) = x² on [1, 3] and find c.
Solution:
- f is continuous on [1, 3] and differentiable on (1, 3) ✓
- f(1) = 1, f(3) = 9
- \(\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4\)
- f'(x) = 2x, so f'(c) = 2c = 4
- c = 2 ∈ (1, 3) ✓
Problem 2: Use Rolle's theorem to prove x³ - 3x + 1 = 0 has at least two roots in (-2, 2).
Solution:
- f(-2) = -8 + 6 + 1 = -1
- f(0) = 1
- f(2) = 8 - 6 + 1 = 3
- By IVT: one root in (-2, 0), another possibility in (0, 2)? f(1) = -1 < 0
- Root in (0, 1) and root in (1, 2)
- Between two roots, by Rolle's theorem, f'(c) = 0 somewhere
- f'(x) = 3x² - 3 = 0 ⟹ x = ±1 (two critical points)
- This confirms at least two roots in (-2, 2)
Problem 3: Show that \(\sin x < x\) for x > 0.
Solution:
- Let f(x) = x - sin x
- f(0) = 0
- f'(x) = 1 - cos x ≥ 0 for all x (equals 0 only when x = 2nπ)
- f is non-decreasing, and strictly increasing except at isolated points
- For x > 0: f(x) > f(0) = 0
- Therefore, x > sin x for x > 0 ✓
Hardest GATE Questions
Topic: Finding the Mean Value Point
Question (GATE 2017 Style): For f(x) = ln x on [1, e], find the value of c guaranteed by MVT.
Solution:
- f(1) = 0, f(e) = 1
- \(\frac{f(e) - f(1)}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1}\)
- f'(x) = 1/x, so f'(c) = 1/c
- \(\frac{1}{c} = \frac{1}{e-1}\)
- c = e - 1 ≈ 1.718 ∈ (1, e) ✓
Why it's hard: Requires careful algebraic manipulation with natural logarithm.
Topic: Cauchy's MVT Application
Question (GATE 2015 Variant): Using Cauchy's MVT, prove that \(\frac{\ln b - \ln a}{b - a} = \frac{1}{c}\) for some c ∈ (a, b) where 0 < a < b.
Solution:
- Let f(x) = ln x, g(x) = x
- f'(x) = 1/x, g'(x) = 1
-
By Cauchy's MVT: $\(\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\)$
-
\[\frac{1/c}{1} = \frac{\ln b - \ln a}{b - a}\]
- Therefore, \(\frac{\ln b - \ln a}{b - a} = \frac{1}{c}\) for some c ∈ (a, b) ✓
Corollary: \(\ln b - \ln a < \frac{b - a}{a}\) (since c > a implies 1/c < 1/a)
Topic: Using MVT to Prove Bounds
Question: Show that for x > 0: \(\frac{x}{1+x} < \ln(1+x) < x\)
Solution: Upper bound (ln(1+x) < x):
- Let f(t) = ln(1+t), apply MVT on [0, x]
- f(x) - f(0) = f'(c) · x for some c ∈ (0, x)
- ln(1+x) = \(\frac{1}{1+c} \cdot x\)
- Since c > 0: \(\frac{1}{1+c} < 1\)
- Therefore, ln(1+x) < x ✓
Lower bound (x/(1+x) < ln(1+x)):
- Since c < x: \(\frac{1}{1+c} > \frac{1}{1+x}\)
- ln(1+x) = \(\frac{x}{1+c} > \frac{x}{1+x}\) ✓