Problem 42 - Calculus

Question

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(g(0)=1\) and \(g(x)\) is increasing for \(x \geq 0,\) then \(f(x)\) is also increasing for \(x \geq 0.\)

Step-by-Step Solution

Answer

True; since \(g(x)\) is increasing and positive for \(x \geq 0\), \(f(x)\) must also be increasing.

Step 1: Understanding the Problem

We are given that \( y = f(x) \) is a solution to the equation \( \frac{dy}{dx} = g(x) \), and we need to determine if the claim that if \( g(0) = 1 \) and \( g(x) \) is increasing for \( x \geq 0 \), then \( f(x) \) is also increasing for \( x \geq 0 \) is true or false.

Step 2: Assessing Derivatives and Function Behavior

The derivative \( \frac{dy}{dx} = g(x) \) implies that the function \( f(x) \) will increase exactly when \( g(x) > 0 \) because this means the slope \( \frac{dy}{dx} > 0 \). Given that \( g(0) = 1 \) and \( g(x) \) is increasing, we need to consider whether these conditions ensure \( g(x) > 0 \) for all \( x \geq 0 \).

Step 3: Checking the Conditions

Since \( g(0) = 1 \) and \( g(x) \) is increasing, it means that for all \( x \geq 0 \), \( g(x) \geq 1 \), which ensures \( g(x) > 0 \). As a result, since \( \frac{dy}{dx} = g(x) \) and \( g(x) \) is greater than or equal to 1, \( f(x) \) must be increasing for all \( x \geq 0 \).

Step 4: Conclusion

Because \( g(x) \) is increasing for \( x \geq 0 \) and initially positive at \( x = 0 \), \( f(x) \) indeed is also increasing for \( x \geq 0 \). Therefore, the statement is true.

Key Concepts

Derivative

The concept of a derivative is central to calculus. It represents the rate at which one quantity changes with respect to another. In simple terms, the derivative of a function at a given point is the slope of the tangent line to the function at that point. Mathematically, if we have a function \( y = f(x) \), the derivative at a point \( x \) is written as \( \frac{dy}{dx} \). This notation communicates the change in \( y \) per unit change in \( x \).

Taking the derivative allows us to understand how a function behaves when \( x \) changes. For example:

• If \( \frac{dy}{dx} > 0 \), the function is increasing; \( y \) grows as \( x \) increases.
• If \( \frac{dy}{dx} < 0 \), the function is decreasing; \( y \) decreases as \( x \) increases.
• If \( \frac{dy}{dx} = 0 \), the function is constant or has a local maximum or minimum.

Understanding derivatives is key to solving differential equations, which describe how functions change in a dynamic context.

Increasing Function

An increasing function is a function where the value of \( f(x) \) goes up as \( x \) increases. In mathematical terms, a function \( f(x) \) is increasing on an interval if for any two numbers \( x_1 < x_2 \), the inequality \( f(x_1) \leq f(x_2) \) holds.

For a function to be increasing, its derivative must be non-negative throughout the given interval. This means that \( \frac{dy}{dx} \geq 0 \) throughout the interval of interest. If \( \frac{dy}{dx} > 0 \) consistently, the function is not only increasing but is strictly increasing, meaning there are no flat segments.

Considering the exercise, since \( g(x) \) is increasing for \( x \geq 0 \) with \( g(0) = 1 \), it implies that \( g(x) \) is always positive and, therefore, the derivative \( \frac{dy}{dx} \) remains positive, and \( f(x) \) is increasing.

Solution to Differential Equation

A differential equation involves an unknown function and its derivatives and is a powerful tool to model real-world phenomena where change is continuous. Solving a differential equation means finding the function or functions that satisfy the equation. In our context, we have the equation \( \frac{dy}{dx} = g(x) \). A solution to this differential equation is any function \( y = f(x) \) for which \( \frac{dy}{dx} = g(x) \) holds for all \( x \) in the function's domain.

To solve such equations, one commonly integrates \( g(x) \) to find \( y = f(x) \). This is because integration is essentially the reverse process of taking a derivative. For example, if \( \frac{dy}{dx} = 2x \), integrating would yield \( y = x^2 + C \), where \( C \) is a constant that can be determined if an initial condition is provided.

In our exercise, with the derivative \( \frac{dy}{dx} = g(x) \) being positive and increasing, its integral \( f(x) \) should reflect this behavior as well, confirming the increase across the interval for \( x \geq 0 \). This comprehensive understanding of solutions helps in confirming or refuting claims about the behavior of functions.