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Chapter 5: Problem 3

Show that both \(f(x):=x\) and \(g(x):=x-1\) are strictly increasing on \(I:=[0,1]\), but that their product \(f g\) is not increasing on \(I\).

Short Answer

Expert verified
Both functions \(f(x):=x\) and \(g(x):=x-1\) are strictly increasing on the interval \(I:=[0,1]\) as their derivatives are positive. However, their product \(f(x)g(x) = x(x - 1)\) is not strictly increasing, as its derivative \(2x - 1\) is not always positive within the interval.

Step by step solution

01

Determining Strictly Increasing Nature of f(x)

A function is considered strictly increasing on an interval if its derivative is always positive therein. Therefore, we should find the derivative of \(f(x)=x\), which is \(f'(x)=1\). Since the derivative of \(f(x)\) is a positive constant, we can conclude that \(f(x)\) is strictly increasing on the interval \(I:=[0,1]\).
02

Determining Strictly Increasing Nature of g(x)

We can repeat the process for \(g(x)=x-1\). The derivative of \(g(x)\), \(g'(x)=1\), works similarly to \(f(x)\) due to the derivative again being a constant positive number. As such, both \(g(x)\) and \(f(x)\) are strictly increasing on \(I\).
03

Prove Non-Increasing Nature of the Product fg(x)

To test the behavior of \(f(x)g(x)\), we first construct the product which gives us \(f(x)g(x) = x(x - 1)\). The derivative of the resulting function is \(f'(x)g(x) + f(x)g'(x) = (1)(x - 1) + (x)(1) = 2x - 1\). On the interval \(I: [0, 1]\), the derivative is not always positive (it is negative for \(x < 0.5\) and positive for \(x > 0.5\)). Therefore, \(f(x)g(x)\) is not strictly increasing on the interval \(I\).

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Most popular questions from this chapter

Let \(I \subseteq \mathbb{R}\) be an interval and let \(f: I \rightarrow \mathbb{R}\) be increasing on \(I\). Suppose that \(c \in I\) is not an endpoint of \(I\). Show that \(f\) is continuous at \(c\) if and only if there exists a sequence \(\left(x_{n}\right)\) in \(I\) such that \(x_{n}c\) for \(n=2,4,6, \cdots\), and such that \(c=\lim \left(x_{n}\right)\) and \(f(c)=\lim \left(f\left(x_{n}\right)\right)\)

If \(g(x):=\sqrt{x}\) for \(x \in[0,1]\), show that there does not exist a constant \(K\) such that \(|g(x)| \leq K|x|\) for all \(x \in[0,1]\). Conclude that the uniformiy continuous \(g\) is not a Lipschitz function on \([0,1]\).

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