Problem 41 - 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 \(g(x)\) is increasing for \(x>0,\) then so is \(f(x).\)

Step-by-Step Solution

Answer

False; increasing \(g(x)\) doesn't imply increasing \(f(x)\), counterexample: \(g(x) = x\).

Step 1: Understand the Statement

The statement suggests that if the derivative of a function \(f(x)\), given by \(g(x)\), is increasing for \(x > 0\), then the function \(f(x)\) itself should also be increasing for \(x>0\). Our goal is to assess the truth of this statement.

Step 2: Analyze the Given Information

The equation \(\frac{dy}{dx} = g(x)\) states that \(g(x)\) is the derivative of \(f(x)\). If \(g(x)\) is increasing, it means its derivative \(g'(x)\) is greater than zero for \(x > 0\). However, \(f(x)\) being increasing depends only on \(g(x) > 0\), not necessarily on \(g(x)\) increasing.

Step 3: Counterexample Consideration

Consider a simple example: let \(g(x) = x\) for \(x > 0\). Its derivative, \(g'(x) = 1\), is not greater than zero since it's constant, but it is indeed not a situation where \(f(x)\) must be increasing. If \(f(x) = \frac{x^3}{3} - x\), then \(g(x) = f'(x) = x^2 - 1\) is increasing for \(x > 1\), but \(f(x)\) is not always increasing because there can be parts where \(f'(x) < 0\).

Step 4: Conclusion

The information that \(g(x)\) is increasing does not imply \(f(x)\) is increasing for \(x > 0\). Instead, \(g(x) > 0\) would be needed for \(f(x)\) to be increasing. Therefore, the statement is false.

Key Concepts

Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. It gives us valuable information about the behavior of the function. If a function is denoted as \(y = f(x)\), its derivative, \(\frac{dy}{dx}\) or \(f'(x)\), reveals how \(y\) changes with small changes in \(x\). The derivative tells us the slope of the tangent line to the function at a particular point.

• If \(f'(x) > 0\), the function is increasing at that point.
• If \(f'(x) < 0\), the function is decreasing.
• If \(f'(x) = 0\), the function could be at a maximum, minimum, or a point of inflection.

Understanding derivatives is critical as they form the basis of many calculus concepts, including optimization and analyzing the behavior of functions over intervals.

Increasing Functions

An increasing function is one where, as you move from left to right along the x-axis, the y-values also increase. In mathematical terms, a function \(f(x)\) is considered to be increasing over an interval if, for every two numbers \(x_1\) and \(x_2\) in the interval, such that \(x_1 < x_2\), the inequality \(f(x_1) \leq f(x_2)\) holds. For a function to be strictly increasing, \(f(x_1) < f(x_2)\) must be true for all \(x_1 < x_2\) within the interval. The concept of increasing functions is closely tied to derivatives. Specifically, if \(f'(x) > 0\) throughout an interval, then \(f(x)\) is increasing on that interval. However, the mere fact that the derivative of a derivative, \(g'(x) > 0\), such as in the original exercise, does not guarantee that \(f(x)\) is increasing, as it only speaks about the behavior of \(g(x)\), not \(f(x)\), directly. This nuanced understanding is critical in analyzing and graphing functions.

Counterexample

A counterexample helps demonstrate that a particular statement or premise is false. In mathematics, simply showing one counterexample is sufficient to disprove a hypothesis. This is a powerful tool since it only takes a single example where the statement does not hold to invalidate it. Consider the claim: "If \(g(x)\) is increasing for \(x > 0\), then so is \(f(x)\)." A counterexample can be given by choosing a function \(f(x)\) where \(g(x) = f'(x)\) is increasing, but \(f(x)\) is not. An instance of this can be when \(g(x) = x\), which has a constant derivative \(g'(x) = 1\). Although \(g(x)\) seems trivially increasing, \(f(x) = \frac{x^3}{3} - x\) has a derivative \(f'(x) = x^2 - 1\) that proves \(f(x)\) is not always increasing as discussed in the original exercise solution. This demonstrates how a clear understanding of derivatives and context is necessary when evaluating mathematical statements.

Differentiation

Differentiation is the core process of finding a derivative. It involves calculating the instantaneous rate of change of a function with respect to one of its variables. In practical terms, this process finds the slope of the curve at any given point, giving an effective way to measure how a function behaves and changes. To differentiate a function, one must:

• Identify the function and its variables.
• Apply differentiation rules such as the power rule, product rule, or chain rule, as applicable.
• Simplify the resulting expression into its most reduced form.

Differentiation is essential for solving many real-world problems, such as finding maxima and minima, determining velocity and acceleration in physics, and solving complex equations. It enables mathematicians and scientists to analyze the dynamic behavior of systems, predict outcomes, and optimize processes across various fields.