Math Problem Statement

. a) Let f: [a,b] Rhe such that f' * (x) >= 0 at each point in (a, b). Prove that f is monotonically. increasing in [a,b].

Solution

To prove that the function f:[a,b]Rf: [a,b] \rightarrow \mathbb{R} is monotonically increasing on [a,b][a, b], given that f(x)0f'(x) \geq 0 for every x(a,b)x \in (a, b), we can follow these steps:

Step-by-Step Proof:

1. Definition of Monotonically Increasing:

A function ff is said to be monotonically increasing on an interval [a,b][a, b] if for every x1,x2[a,b]x_1, x_2 \in [a, b] such that x1<x2x_1 < x_2, we have f(x1)f(x2)f(x_1) \leq f(x_2).

Thus, we need to show that f(x1)f(x2)f(x_1) \leq f(x_2) whenever x1<x2x_1 < x_2, using the fact that f(x)0f'(x) \geq 0 for all x(a,b)x \in (a, b).

2. Use of the Mean Value Theorem:

The Mean Value Theorem (MVT) states that if a function ff is continuous on a closed interval [x1,x2][x_1, x_2] and differentiable on the open interval (x1,x2)(x_1, x_2), then there exists some c(x1,x2)c \in (x_1, x_2) such that: f(c)=f(x2)f(x1)x2x1.f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.

Since f(x)0f'(x) \geq 0 for all x(a,b)x \in (a, b), we have f(c)0f'(c) \geq 0. Thus, applying the Mean Value Theorem for ff on [x1,x2][x_1, x_2], we get: f(x2)f(x1)x2x10.\frac{f(x_2) - f(x_1)}{x_2 - x_1} \geq 0.

This implies that: f(x2)f(x1)0,f(x_2) - f(x_1) \geq 0, or equivalently: f(x1)f(x2).f(x_1) \leq f(x_2).

3. Conclusion:

Thus, we have shown that f(x1)f(x2)f(x_1) \leq f(x_2) for all x1<x2x_1 < x_2 in [a,b][a, b]. Therefore, ff is monotonically increasing on [a,b][a, b].

Conclusion:

The function ff is monotonically increasing on [a,b][a, b], as required, because f(x)0f'(x) \geq 0 for all x(a,b)x \in (a, b), and the Mean Value Theorem ensures that the function satisfies the condition of monotonicity.


Would you like further details or have any questions?

Here are 5 related questions to explore the concept further:

  1. How does the Mean Value Theorem relate to differentiability?
  2. Can you have a function where f(x)=0f'(x) = 0 everywhere but it is not constant?
  3. If f(x)0f'(x) \leq 0, what can be said about the function's monotonicity?
  4. What happens if f(x)f'(x) is not defined at some points in (a,b)(a, b)?
  5. Can a function be increasing but not differentiable at certain points?

Tip: When applying the Mean Value Theorem, remember that the function must be continuous on the closed interval and differentiable on the open interval for the theorem to hold.

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Math Problem Analysis

Mathematical Concepts

Calculus
Monotonicity
Differentiation

Formulas

f'(x) ≥ 0 implies f is monotonically increasing
Mean Value Theorem: f'(c) = (f(x2) - f(x1)) / (x2 - x1)

Theorems

Mean Value Theorem
Monotonicity Theorem

Suitable Grade Level

College Calculus (Undergraduate)