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Chapter 3: Problem 69

Increasing Functions Is the product of two increasing functions always increasing? Explain.

Short Answer

Expert verified
No, the product of two increasing functions is not always increasing. It depends on whether the functions can take on negative values. If both functions are always non-negative, then their product will be increasing, otherwise not necessarily.

Step by step solution

01

Understanding the Definition of an Increasing Function

A function \( f(x) \) is said to be increasing on an interval if for any two numbers \( x_1 \) and \( x_2 \) in the interval, \( x_1 < x_2 \) implies \( f(x_1) < f(x_2) \). In other words, as \( x \) increases, \( f(x) \) also increases.
02

Considering the Product of Two Increasing Functions

Let's consider two increasing functions \( f(x) \) and \( g(x) \). The product function is \( h(x) = f(x) g(x) \). To determine if \( h(x) \) is always increasing, we need to check if \( h(x_1) < h(x_2) \) whenever \( x_1 < x_2 \). Since \( f \) and \( g \) are both increasing functions, anywhere \( x_1 < x_2 \), it's true that \( f(x_1) < f(x_2) \) and \( g(x_1) < g(x_2) \). However, if the functions \( f \) and \( g \) can take on negative values, then the product \( h(x) \) can decrease as \( x \) increases. For example, consider \( f(x) = x \) and \( g(x) = -x \) for \( x \) in the interval [-1,1]. Both \( f \) and \( g \) are increasing functions, but the product function \( h(x) = x*(-x) = -x^2 \) is decreasing.
03

Concluding the Analysis

Thus, it is concluded that the product of two increasing functions is not always increasing. It depends on whether the functions can take on negative values. In the case that both functions are increasing and are always non-negative, then the product will be an increasing function, otherwise not necessarily.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product of Functions
When we talk about the product of functions, we're considering the multiplication of two different functions to create a third new function. If you have two functions, say \( f(x) \) and \( g(x) \), their product is \( h(x) = f(x) \cdot g(x) \). This means for each value of \( x \), \( h(x) \) is the result of multiplying the corresponding outputs of \( f(x) \) and \( g(x) \).

This new function \( h(x) \) derives its characteristics from both \( f \) and \( g \). Whether it increases, decreases, or remains constant can depend on various factors, such as the individual behavior of \( f(x) \) and \( g(x) \).

One might expect that multiplying functions with similar behaviors (e.g., both increasing) might give a predictable result, but this is not always the case. The sign and magnitude of the functions play a crucial role, especially when negative values are involved.
Negative Values
Negative values in a function can strongly influence the way a product function behaves. Suppose both \( f(x) \) and \( g(x) \) are increasing functions. It seems intuitive to think their product \( h(x) = f(x)g(x) \) would also be increasing.

However, this changes if one or both of these functions take on negative values. Multiplying a positive number with a negative one results in a negative number. Similarly, multiplying two negative numbers results in a positive number. This switch in sign can make a product function behave differently than expected.

For instance, let's consider \( f(x) = x \) and \( g(x) = -x \) over the interval \([-1, 1]\). Individually, both functions are increasing across this range. However, their product, \( h(x) = -x^2 \), behaves differently. Instead of increasing, \( h(x) \) actually decreases, demonstrating the powerful impact of negative values.
Non-negative Functions
Non-negative functions are those which do not take on any negative values; they are always zero or positive. When we consider two such functions, both being non-negative and increasing, their product is often easier to predict.

With both functions remaining on the positive side of the number line, the product function will indeed be increasing. Why? Because multiplying two positives gives a positive, and their increasing nature contributes to the overall increase in the product function.

For example, if \( f(x) = x + 1 \) and \( g(x) = x + 2 \) over a range where \( x > 0 \), both are non-negative and increase as \( x \) increases. Their product, \( h(x) = (x + 1)(x + 2) \), increases smoothly without any complicating sign changes.

This illustrates that non-negative functions maintain a consistent growth pattern when multiplied, highlighting the importance of non-negative values when analyzing the behavior of product functions.

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Most popular questions from this chapter

Writing In Exercises \(33-36\) , explain why the Mean Value Theorem does not apply to the function \(f\) on the interval \([0,6]\) . \(f(x)=|x-3|\)

Comparing \(\Delta y\) and \(d y\) In Exercises \(13-18\) use the information to find and compare \(\Delta y\) and \(d y\) . $$\begin{array}{ll}{\text { Function }} & {x \text { -Value }} \\ {y=0.5 x^{3}} & {x=1}\end{array} \quad \begin{array}{ll}{\text { Differential of } x} \\\ {\Delta x=d x=0.1}\end{array}$$

Using Newton's Method In Exercises \(7-16,\) use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations differ by less than 0.001. Then find the zero(s) using a graphing utility and compare the results. \(f(x)=2-x^{3}\)

Maximum Area Twenty feet of wire is to be used to form two figures. In each of the following cases, how much wire should be used for each figure so that the total enclosed area is maximum? \begin{equation} \begin{array}{l}{\text { (a) Equilateral triangle and square }} \\ {\text { (b) Square and regular pentagon }} \\ {\text { (c) Regular pentagon and regular hexagon }} \\ {\text { (d) Regular hexagon and circle }}\end{array} \end{equation} What can you conclude from this pattern? \(\\{\)Hint\(:\) The area of a regular polygon with \(n\) sides of length \(x\) is \(A=(n / 4)[\cot (\pi / n)] x^{2} . \\}\)

Rolle's Theorem Let \(f\) be continuous on \([a, b]\) and differentiable on \((a, b) .\) Also, suppose that \(f(a)=f(b)\) and that \(c\) is a real number in the interval \((a, b)\) such that \(f^{\prime}(c)=0 .\) Find an interval for the function \(g\) over which Rolle's Theorem can be applied, and find the corresponding critical number of \(g\) , where \(k\) is a constant. (a) $$g(x)=f(x)+k$$ (b) $$g(x)=f(x-k)$$ (c) $$g(x)=f(k x)$$
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