Problem 40 - 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 all \(x,\) then the graph of \(f\) is concave up for all \(x.\)

Step-by-Step Solution

Answer

False. An increasing \(g(x)\) does not ensure \(f(x)\) is concave up.

Step 1: Understanding Function's Derivative

Given that \(y = f(x)\) is a solution to the equation \(\frac{dy}{dx} = g(x)\), this means that \(f'(x) = g(x)\). We are asked whether an increasing \(g(x)\) implies that \(f(x)\) is concave up.

Step 2: Relating Concavity to Second Derivative

A function \(f(x)\) is concave up when its second derivative \(f''(x) > 0\). To explore this condition, we need \(f''(x) = g'(x)\) to be positive.

Step 3: Checking the Statement's Truth

Since \(g(x)\) is increasing, it implies that \(g'(x) \geq 0\) for all \(x\). However, for a function to be strictly increasing, \(g'(x)\) must be greater than zero in some neighborhood. Therefore, even though \(g(x)\) is increasing, it doesn't necessarily follow that \(g'(x) > 0\) for all \(x\).

Step 4: Providing Counterexample

Consider \(g(x) = x^2\), which is increasing for all \(x \geq 0\) since \(g'(x) = 2x \geq 0\) for \(x \geq 0\). However, \(g'(x) = 0\) at \(x = 0\), meaning \(f''(x)=0\) at that point, violating the condition needed for \(f(x)\) to be concave up.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point. This idea is pivotal when analyzing how a function behaves. When we talk about the derivative of a function, we're looking at the slope of the tangent line to the function's graph at any point. This slope tells us how steeply the function is increasing or decreasing.
In the original exercise, we explored a situation where the derivative of a function is a separate function, noted as \(g(x)\). If \(y = f(x)\) solves \(\frac{dy}{dx} = g(x)\), it implies that \(f'(x) = g(x)\). The task was to determine if \(g(x)\) being an increasing function automatically makes \(f(x)\) concave up.
This introduces the concept of the second derivative, which helps us evaluate concavity.

Concavity

Concavity tells us how the slope of the tangent line changes as you move along the curve of a function. When a function is concave up, it means that as you move along the curve from left to right, the curve opens upwards, like a smile. Mathematically, this is determined by the second derivative of the function. If the second derivative, \(f''(x)\), is positive, the function is concave up.
In our context, after finding \(f'(x) = g(x)\), the second derivative is \(f''(x) = g'(x)\). For the function \(f(x)\) to be concave up, we need \(g'(x) > 0\). However, simply having \(g(x)\) as an increasing function doesn't mean \(g'(x) > 0\) everywhere; it might be zero at some points. Thus, an increasing \(g(x)\) doesn't guarantee \(f(x)\) is concave up everywhere. Ensure to check \(g'(x)\) before concluding about the concavity of \(f(x)\).

Increasing Functions

When a function is increasing, it means that as you move from left to right along the x-axis, the y-values (output of the function) also increase. Formally, a function \(g(x)\) is increasing on an interval if \(g'(x) \geq 0\) throughout that interval.
In the original exercise's context, we assumed that \(g(x)\) was an increasing function. However, just because \(g(x)\) is increasing doesn't mean \(g'(x) > 0\) everywhere. A function can be increasing and have regions where its derivative equals zero. For example, \(g(x) = x^2\) is increasing for \(x \geq 0\), but its derivative \(g'(x) = 2x\) equals zero when \(x = 0\).
Therefore, while \(g(x)\) increasing suggests some structure in \(f(x)\), we cannot solely rely on this to determine concavity, which requires a positive second derivative.