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In Problems 1-4, show that each function is continuous at the given value. $$ f(x)=2 x, c=1 / 2 $$

Short Answer

Expert verified
The function \( f(x) = 2x \) is continuous at \( x = \frac{1}{2} \) because the limit and function value at this point are equal.

Step by step solution

01

Definition of Continuity

To show that the function is continuous at a given point, we need to verify that for any point \( c \), the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \). Specifically, for the function \( f(x) = 2x \) and \( c = \frac{1}{2} \), we need to verify that \( \lim_{x \to \frac{1}{2}} f(x) = f\left(\frac{1}{2}\right) \).
02

Evaluate the Function at c

Calculate \( f(c) \) using the function \( f(x) = 2x \). Substitute \( c = \frac{1}{2} \) into the function: \[ f\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} = 1. \]
03

Evaluate the Limit as x Approaches c

Calculate the limit of \( f(x) = 2x \) as \( x \rightarrow \frac{1}{2} \). Substitute \( x = \frac{1}{2} \) into the expression: \[ \lim_{x \to \frac{1}{2}} 2x = 2 \cdot \lim_{x \to \frac{1}{2}} x = 2 \cdot \frac{1}{2} = 1. \]
04

Compare the Limit and the Function Value

We found that \( \lim_{x \to \frac{1}{2}} f(x) = 1 \) and \( f\left(\frac{1}{2}\right) = 1 \). Since these values are equal, the function is continuous at \( x = \frac{1}{2} \).

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Key Concepts

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

Understanding Limits
Limits are a fundamental concept in calculus that help us understand the behavior of functions as they approach a certain point.
The idea is to see what value a function approaches as the input, usually denoted as \( x \), gets closer and closer to a particular value, denoted as \( c \).
For instance, with the function \( f(x) = 2x \), we are interested in understanding what happens as \( x \) approaches \( \frac{1}{2} \).
To find the limit, you analyze the behavior of \( f(x) \) for values of \( x \) that are very close to, but not equal to, \( c \). For the given function, as \( x \) approaches \( \frac{1}{2} \), the value of \( 2x \) gets closer to \( 1 \). This means \( \lim_{x \to \frac{1}{2}} 2x = 1 \).
  • Limits let us reach a conclusion about function values that are too close to discern by direct substitution.
  • They form the basis for calculus operations, allowing us to understand changes in functions' behaviors.
Function Evaluation
Function evaluation is a straightforward process that involves determining the output of a function for a given input.
When assessing function evaluation at a particular point, like \( f\left(\frac{1}{2}\right) \) for the function \( f(x) = 2x \), you simply substitute the given value of \( x \) into the function equation.
In this example, replacing \( x \) with \( \frac{1}{2} \) results in \( 2 \times \frac{1}{2} = 1 \). The output \( f\left(\frac{1}{2}\right) = 1 \) signifies the value of the function when \( x \) is \( \frac{1}{2} \).
  • Function evaluation helps us determine exact values, necessary for validating continuity and limits.
  • It's pivotal in connecting function behavior with theoretical values calculated using limits.
Continuity at a Point
Continuity at a particular point means that a function behaves nicely at that point.
Specifically, it implies no jumps, breaks, or gaps in the function graph at that location.
This can be confirmed if three conditions are met:
  • The function must be defined at the point. In our example, \( f\left(\frac{1}{2}\right) = 1 \) confirms this.
  • The limit must exist as \( x \) approaches the point. As established, \( \lim_{x \to \frac{1}{2}} 2x = 1 \) meets this criterion.
  • The limit and the function value at the point must be equal. Here, both equal 1.
In the context of the function \( f(x) = 2x \) at \( x = \frac{1}{2} \), all conditions are satisfied. Thus, the function is continuous at this point, which means there is a smooth transition without any abrupt changes or undefined behavior when \( x = \frac{1}{2} \). Continuity is crucial for the stability and predictability of functions, ensuring reliable real-world applications.

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