Chapter 2: Problem 64
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.
Short Answer
Expert verified
If \(f(c) > 0\) and \(f\) is continuous at \(c\), there is a \(\delta > 0\) so \(f(x) > 0\) for \(|x - c| < \delta\).
Step by step solution
01
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 \(|f(x) - f(c)| < \epsilon\).
02
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.
03
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 \(|f(x) - f(c)| < \frac{f(c)}{2}\).
04
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 \(f(c) - \frac{f(c)}{2} < f(x) < f(c) + \frac{f(c)}{2}\).
05
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 \(\frac{f(c)}{2} > 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( f(c)-\epsilon < f(x) < f(c)+\epsilon \).
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 \( c \).
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.
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.
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 \( \left(f(c) - \frac{f(c)}{2}, f(c) + \frac{f(c)}{2}\right) \).
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.