Problem 61 - Calculus with Concepts in Calculus

Question

Let \(f\) and \(g\) be continuous on \([a, \infty)\) and differentiable on \((a, \infty)\). Assume that \(f(a) \geq g(a)\) and \(f^{\prime}(x)>g^{\prime}(x)\) for all \(x>a\). Prove that \(f(x)>g(x)\) for any \(x>a\).

Step-by-Step Solution

Answer

Since \(f'(x) > g'(x)\) for all \(x > a\), \(f(x) > g(x)\) for any \(x > a\).

Step 1: Understand Initial Conditions

We are given that \(f\) and \(g\) are continuous on \([a, \infty)\) and differentiable on \((a, \infty)\). The conditions \(f(a) \geq g(a)\) and \(f'(x) > g'(x)\) for all \(x > a\) need to be used to show that \(f(x) > g(x)\) for any \(x > a\).

Step 2: Define a New Function

Consider the function \( h(x) = f(x) - g(x) \). Our objective is to show that \( h(x) > 0 \) for \( x > a \). Initially, we have \( h(a) = f(a) - g(a) \geq 0 \).

Step 3: Analyze the Derivative of the New Function

Determine the derivative \( h'(x) = f'(x) - g'(x) \). Given that \( f'(x) > g'(x) \), it follows that \( h'(x) > 0 \) for all \( x > a \).

Step 4: Inferring the Increasing Nature of the Function

Since \( h'(x) > 0 \) for all \( x > a \), the function \( h(x) \) is increasing on \((a, \infty)\). This implies that \( h(x) > h(a) \geq 0 \) for all \( x > a \).

Step 5: Conclusion

Given \( h(x) = f(x) - g(x) \) is increasing and \( h(a) \geq 0 \), we conclude that \( h(x) > 0 \) for all \( x > a \). Thus, \( f(x) > g(x) \) for any \( x > a \).

Key Concepts

Continuous Functions

Continuous functions are fundamental concepts in calculus. They ensure a smooth flow without breaks or jumps in the graph of a function.
Imagine drawing a curve on paper without lifting your pencil. That's a visual representation of continuity.

Mathematically, a function \( f(x) \) is continuous on an interval \([a, \infty)\) if:\[ \lim_{x \to c} f(x) = f(c) \] for every point \( c \) in the interval.
This means for any small change in \( x \), the change in \( f(x) \) is equally small.

• This characteristic helps in analyzing functions over an interval.
• Continuity also ensures that limits exist at every point within the interval.
• For our problem, both functions \( f \) and \( g \) are continuous, which forms the base for further deductions.

Knowing that the functions are continuous supports conclusions about their behavior, ensuring no abrupt changes which would otherwise invalidate derivative comparisons.

Derivatives

Derivatives express how functions change at any given point. They are powerful tools in calculus to investigate function behavior and trends.
The derivative of a function \( f(x) \), denoted \( f'(x) \), represents the rate of change of \( f(x) \) with respect to \( x \).

In our problem, we know that \( f'(x) > g'(x) \) for all \( x > a \). This means that the slope, or the rate of change of \( f(x) \), is consistently greater than that of \( g(x) \).

• A positive derivative \( f'(x) \) indicates that \( f(x) \) is increasing.
• If \( f'(x) > g'(x) \), then \( f(x) \) grows faster than \( g(x) \).
• This supports the conclusion that \( f(x) \) surpasses \( g(x) \) as \( x \) increases.

Derivatives not only describe the current behavior of functions but also allow us to predict future behavior on the same interval.

Mathematical Proofs

Mathematical proofs are logical arguments verifying the truth of a theorem or statement. They use known facts, definitions, properties, and logical reasoning.
Proofs are crucial for confirming results, ensuring that conclusions hold under given conditions.

In this exercise, we utilized a structured proof to show that \( f(x) > g(x) \) for \( x > a \). We defined \( h(x) = f(x) - g(x) \) and showed that:\[ h(x) > 0 \] for \( x > a \) using follow-up steps:

• Since \( f(a) \geq g(a) \), then \( h(a) \geq 0 \).
• With \( h'(x) = f'(x) - g'(x) > 0 \), \( h(x) \) is increasing, thus \( h(x) > h(a) \geq 0 \).

The above steps reinforce the conditions are met and ensure the validity of the statements about \( f(x) \) and \( g(x) \).
Proofs help confirm that our analysis is comprehensive and correct, grounding conclusions in solid mathematical reasoning.