Math Problem Statement

Prove a function to be increasing using lmvt

Solution

To prove that a function is increasing using Lagrange's Mean Value Theorem (LMVT), we need to go through the following steps:

Lagrange's Mean Value Theorem (LMVT) Statement:

LMVT states that if a function f(x)f(x) is:

  1. Continuous on a closed interval [a,b][a, b],
  2. Differentiable on the open interval (a,b)(a, b),

then there exists at least one point c(a,b)c \in (a, b) such that:

f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}

Now, if we can show that f(x)f'(x) is positive for all x(a,b)x \in (a, b), it implies that f(b)>f(a)f(b) > f(a) for all a<ba < b, meaning the function is increasing.

Steps to prove a function is increasing using LMVT:

  1. Check Continuity and Differentiability:

    • Ensure the function f(x)f(x) is continuous on the closed interval [a,b][a, b].
    • Ensure the function f(x)f(x) is differentiable on the open interval (a,b)(a, b).
  2. Apply LMVT: By LMVT, for some c(a,b)c \in (a, b), we have: f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a} The sign of f(c)f'(c) will indicate the behavior of f(x)f(x).

  3. Check the Derivative:

    • If f(x)>0f'(x) > 0 for all x(a,b)x \in (a, b), then the slope of the secant line f(b)f(a)ba\frac{f(b) - f(a)}{b - a} will be positive for all a<ba < b, proving that f(x)f(x) is strictly increasing on [a,b][a, b].

Example:

Let's consider the function f(x)=x2f(x) = x^2, and prove it is increasing on the interval [1,3][1, 3].

  1. Continuity and Differentiability: f(x)=x2f(x) = x^2 is continuous and differentiable on all real numbers, so it's continuous on [1,3][1, 3] and differentiable on (1,3)(1, 3).

  2. Apply LMVT: By LMVT, there exists some c(1,3)c \in (1, 3) such that: f(c)=f(3)f(1)31=912=4f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4 So, for some c(1,3)c \in (1, 3), f(c)=4f'(c) = 4.

  3. Check the Derivative: The derivative of f(x)=x2f(x) = x^2 is f(x)=2xf'(x) = 2x, which is positive for all x>0x > 0. Therefore, on the interval [1,3][1, 3], f(x)>0f'(x) > 0, which means f(x)f(x) is increasing on [1,3][1, 3].

Thus, we have proved using LMVT that f(x)=x2f(x) = x^2 is increasing on the interval [1,3][1, 3].


Would you like more details on this process or another example?

Here are some related questions:

  1. How does LMVT relate to the concept of monotonicity in functions?
  2. Can LMVT be used to prove a function is decreasing?
  3. What are the conditions for a function to be strictly increasing?
  4. How does Rolle's Theorem differ from LMVT?
  5. How do you prove a function is increasing without using LMVT?

Tip: To check whether a function is increasing or decreasing, always analyze the sign of its derivative.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Monotonicity

Formulas

f'(c) = (f(b) - f(a)) / (b - a)
f'(x) > 0 implies f(x) is increasing

Theorems

Lagrange's Mean Value Theorem (LMVT)

Suitable Grade Level

Grades 11-12