Problem 64 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that if \(f\) is continuous at \(c\) and \(f(c)>0\) there is an interval \((c-\delta, c+\delta)\) such that \(f(x)>0\) on this interval.

Step-by-Step Solution

Answer

If \(f(c) > 0\) and \(f\) is continuous at \(c\), there is a \(\delta > 0\) so \(f(x) > 0\) for \(|x - c| < \delta\).

Step 1: Understanding Continuity

Since \(f\) is continuous at point \(c\), we know that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x - c| < \delta\), then \(|f(x) - f(c)| < \epsilon\).

Step 2: Setting the Target for \(\epsilon\)

Given that \(f(c) > 0\), we need to find an interval where \(f(x) > 0\) as well. Let's choose \(\epsilon = \frac{f(c)}{2}\). This specific choice of \(\epsilon\) helps ensure that \(f(x)\) stays positive.

Step 3: Applying the Epsilon-Delta Definition

With \(\epsilon = \frac{f(c)}{2}\), by continuity, there exists a \(\delta > 0\) such that for \(|x - c| < \delta\), we have \(|f(x) - f(c)| < \frac{f(c)}{2}\).

Step 4: Ensuring Positivity of Values

The inequality \(|f(x) - f(c)| < \frac{f(c)}{2}\) implies \(-\frac{f(c)}{2} < f(x) - f(c) < \frac{f(c)}{2}\). Adding \(f(c)\) throughout, we get \(f(c) - \frac{f(c)}{2} < f(x) < f(c) + \frac{f(c)}{2}\).

Step 5: Simplifying the Inequality

The above inequality simplifies to \(\frac{f(c)}{2} < f(x) < \frac{3f(c)}{2}\). Thus, \(f(x) > 0\) on \((c-\delta, c+\delta)\) since \(\frac{f(c)}{2} > 0\).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a fundamental concept in calculus used to define continuity rigorously. It helps us understand how small changes in the input of a function lead to small changes in the output.
Imagine having a function \( f \) that is continuous at a specific point \( c \). According to the epsilon-delta definition, for every positive number \( \epsilon \) (epsilon represents a tiny amount), there is a corresponding positive number \( \delta \) (delta is also a tiny amount) such that if the input \( x \) is within the range \( c-\delta < x < c+\delta \), then the output \( f(x) \) will be within the range \( f(c)-\epsilon < f(x) < f(c)+\epsilon \).

• \( \epsilon \) and \( \delta \) illustrate how changes in \( x \) (input) lead to bounded changes in \( f(x) \) (output).
• Choosing \( \epsilon \) intelligently helps us preserve properties like positivity around the point \( c \).

This definition provides a precise way to capture the idea of a function being smooth without any sudden jumps or breaks at a point.

Continuous Function

A continuous function is a type of function that, intuitively, you can draw without lifting your pencil from the paper. This means that small changes in the input \( x \) result in small changes in the output \( f(x) \).
For a function \( f \) to be continuous at a point \( c \), the limit of \( f(x) \) as \( x \) approaches \( c \) must be equal to \( f(c) \). This is where the epsilon-delta definition plays a crucial role.

• If \(|x - c| < \delta\), then the values \( f(x) \) are close to \( f(c) \).
• This closeness is ensured by choosing \( \epsilon \) and \( \delta \) appropriately, as discussed earlier.

Continuity at a point ensures that the function behaves predictably and smoothly around that point, avoiding any abrupt changes or gaps. This concept is vital in numerous mathematical contexts and applications, allowing us to make reliable predictions about function behavior in small intervals around specific points.

Positivity of Functions

The positivity of functions describes when a function yields positive output values. In the context of a continuous function, particularly at a point \( c \) where \( f(c) > 0 \), we want to ensure that nearby values of \( x \) will also give positive \( f(x) \).
Given the scenario where \( f(c) > 0 \), using the epsilon-delta definition, we can choose \( \epsilon = \frac{f(c)}{2} \). This choice of \( \epsilon \) ensures that \( f(x) \) remains within a certain window around \( f(c) \), specifically within \( \left(f(c) - \frac{f(c)}{2}, f(c) + \frac{f(c)}{2}\right) \).

• This translates to \( \frac{f(c)}{2} < f(x) < \frac{3f(c)}{2} \).
• Since \( \frac{f(c)}{2} > 0 \), \( f(x) \) is positive within this interval.

Therefore, around the point \( c \), within a suitably chosen interval \( (c - \delta, c + \delta) \), the function remains positive. This ensures the function does not unexpectedly dip below zero, maintaining a consistently positive value in that neighborhood.