/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Let \(I\) be an interval and let... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 13

Let \(I\) be an interval and let \(f: I \rightarrow \mathbb{R}\) be differentiable on \(I\). Show that if \(f^{\prime}\) is positive on \(I\), then \(f\) is strictly increasing on \(I\).

Short Answer

Expert verified
The function \(f\) is strictly increasing on the interval \(I\) as for any \(x < y\), \(f(x) < f(y)\).

Step by step solution

01

Define Strictly Increasing

Firstly, let's define what it means for a function to be strictly increasing. A function \(f\) is strictly increasing on an interval \(I\) if for any \(x, y\) in \(I\), if \(x < y\) then \(f(x) < f(y)\). This definition will be used in our proof.
02

Apply Mean Value Theorem

The Mean Value Theorem states that if a function is continuous on the closed interval \([a, b]\), and differentiable on the open interval \((a, b)\), then there exists a \(c\) in \((a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}\] Let's apply this theorem to the function \(f\) on the interval \([x, y]\), such that \(x < y\). Therefore, since \(f\) is assumed to be differentiable on \(I\), there is a number \(c\) in \((x, y)\) such that \[f'(c) = \frac{f(y) - f(x)}{y - x}\]
03

Concluding the Proof

As given, the derivative \(f'(c)\) is positive for all \(c\) in \(I\). So \[f'(c) = \frac{f(y) - f(x)}{y - x} > 0\] which implies that \[f(y) > f(x)\] Therefore, we have shown that for any \(x < y\) in the interval \(I\), \(f(x) < f(y)\), so according to our definition, \(f\) is strictly increasing on \(I\).

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