Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that if \(f(x) \geq 0\) and \(f^{\prime}(x) \geq 0\) on \(I\), then \(f^{2}\) is nondecreasing on \(I\).

Step-by-Step Solution

Answer

Since \( g'(x) = 2f(x)f'(x) \geq 0 \), \( f^2(x) \) is nondecreasing on \( I \).

Step 1: Understand the Given Conditions

We are given that a function \( f(x) \) is non-negative (\( f(x) \geq 0 \)) and its derivative is also non-negative (\( f'(x) \geq 0 \)) on an interval \( I \). We need to prove that the square of the function, \( f^2(x) \), is a nondecreasing function on \( I \).

Step 2: Utilize the Derivative of the Square Function

To prove that \( f^2 \) is nondecreasing, it suffices to show that its derivative is non-negative. Calculate the derivative of \( f^2(x) \) using the chain rule. Set \( g(x) = f^2(x) \), and thus \( g'(x) = 2f(x)f'(x) \).

Step 3: Analyze the Derivative Sign

Since \( f(x) \geq 0 \) and \( f'(x) \geq 0 \), and because these are multiplied by 2 in \( g'(x) = 2f(x)f'(x) \), it follows that \( g'(x) \geq 0 \). A non-negative derivative indicates that \( f^2(x) \) is nondecreasing on the interval \( I \).

Step 4: Conclusion

With \( g'(x) = 2f(x)f'(x) \geq 0 \) for all \( x \) in \( I \), this confirms that \( f^2(x) \) is nondecreasing. Since both \( f(x) \) and \( f'(x) \) are non-negative, their product is non-negative, ensuring \( f^2 \) doesn't decrease.

Key Concepts

Non-decreasing Functions

A function is termed non-decreasing over an interval if it doesn't fall as you move from left to right along the x-axis. In other words, the function maintains its value or increases as you progress further on the graph. This property is very useful in real-life scenarios, like predicting that a particular quantity will either stay steady or inflate over time.

• For a function to be non-decreasing, its derivative must be zero or positive across the interval. That means the slope of the tangent to the function isn't negative.
• A non-negative slope implies that there are no downward trends on the function. Thus, ensuring a steady or increasing pattern.

In the exercise, we confirmed the non-decreasing nature of the function by examining the derivative of the squared function. This derivative check is vital because it validates whether the function behaves as predicted over the specified span.

Derivative Analysis

Analyzing derivatives is a core aspect of calculus, providing insights into how a function is changing at any point. Specifically, for any function, its first derivative gives the rate of change or the 'slope' of the function. Understanding derivatives helps in determining the behavior of functions concerning phenomena like stability or growth.
When considering whether a function is non-decreasing, observing its first derivative is essential.

• If the derivative is positive, the function is rising.
• If it's zero, the function is flat or unchanging.
• If negative, the function is decreasing.

In the context of the proof, we focused on the derivative of the squared function. By finding that the derivative, given as \( g'(x) = 2f(x)f'(x) \), remained non-negative, we established that the whole function was non-decreasing. Derivative analysis thus served as the cornerstone of our understanding and final conclusion in this proof.

Mathematical Proof Techniques

Mathematical proofs are a way to rigorously demonstrate that a statement is true using logic and previously established facts. Different techniques are used depending on the context, but common methods include direct proof, contradiction, and induction.
In our exercise, the proof used was a straightforward direct proof. The steps involved were:

• First, verify the provided conditions and understand their implications.
• Compute relevant derivatives (using chain rules if applicable) to guide the analysis.
• Demonstrate that the conditions ensure the desired behavior—namely, that the function is non-decreasing.

The effectiveness of this proof relied heavily on solid knowledge of derivatives. Logical connections between non-decreasing functions and non-negative derivatives were employed to reach a conclusion. By following this step-by-step framework, the proof validated the non-decreasing nature of our function \( f^2 \) efficiently and accurately.