Chapter 1: Problem 52
Show that the indicated implication is true $$ |x+4|<\frac{\varepsilon}{2} \Rightarrow|2 x+8|<\varepsilon $$
Short Answer
Expert verified
The implication is true, as \(|x+4|<\frac{\varepsilon}{2}\) ensures \(|2x+8|<\varepsilon\).
Step by step solution
01
Understand the Implication
We need to show that whenever the condition \(|x+4|<\frac{\varepsilon}{2}\) is true, the result \(|2x+8|<\varepsilon\) must also be true. This is what is meant by the implication \(|x+4|<\frac{\varepsilon}{2} \Rightarrow |2x+8|<\varepsilon\).
02
Express the Result in Terms of the Condition
Observe that \(2x+8\) can be rewritten as \(2(x+4)\). Hence, \(|2x+8|=|2(x+4)|=2|x+4|\). This step simplifies the problem, allowing us to directly connect the absolute terms.
03
Use the Given Condition in Simplified Result
Since the condition \(|x+4|<\frac{\varepsilon}{2}\) is given, substitute it into our simplified result: \\[|2(x+4)| = 2|x+4| < 2 \times \frac{\varepsilon}{2} = \varepsilon\]
04
Conclusion of the Implication
Since we have shown that if \(|x+4|<\frac{\varepsilon}{2}\), it directly leads to \(|2x+8|=2|x+4|<\varepsilon\), we conclude that the implication is true. Thus, whenever \(|x+4|<\frac{\varepsilon}{2}\), \(|2x+8|<\varepsilon\) will also be satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Values in Context
Understanding absolute values is crucial in grasping many calculus problems. The absolute value of a number is its distance from zero on the real number line, regardless of direction. So, when we write \(|x+4|\), it represents how far \(x+4\) is from zero, ensuring a non-negative expression.
The absolute value concept is applied in inequalities, as seen in the exercise, to provide solutions that hold true regardless of whether the expression within is positive or negative.
When dealing with conditions like \(|x+4|<\frac{\varepsilon}{2}\), it indicates constraints on how small the expression \(x+4\) can be. This tells us that while the precise value of \(x\) might vary, its deviation from -4 is limited to being less than half of \(\varepsilon\). Understanding this helps when linking conditions in implications.
The absolute value concept is applied in inequalities, as seen in the exercise, to provide solutions that hold true regardless of whether the expression within is positive or negative.
When dealing with conditions like \(|x+4|<\frac{\varepsilon}{2}\), it indicates constraints on how small the expression \(x+4\) can be. This tells us that while the precise value of \(x\) might vary, its deviation from -4 is limited to being less than half of \(\varepsilon\). Understanding this helps when linking conditions in implications.
Explaining the Epsilon-Delta Definition
The epsilon-delta definition plays a fundamental role in understanding limits, especially in the context of calculus. It is a formal way of describing how a function behaves as the input approaches a certain value.
In simple terms, for the function \(f(x)\) and limit \(L\), the definition expresses that for every tiny number \(\varepsilon > 0\), there exists another tiny number \(\delta > 0\) such that if \(|x - c| < \delta\), then \(|f(x) - L| < \varepsilon\).
This provides a precise way of defining what it means for \(f(x)\) to get arbitrarily close to \(L\) as \(x\) gets arbitrarily close to \(c\). In the given exercise, the \(\varepsilon\) is used to dictate how close \(2x + 8\) must be to zero, establishing a relationship with \(|x+4|<\frac{\varepsilon}{2}\) that maintains these conditions.
In simple terms, for the function \(f(x)\) and limit \(L\), the definition expresses that for every tiny number \(\varepsilon > 0\), there exists another tiny number \(\delta > 0\) such that if \(|x - c| < \delta\), then \(|f(x) - L| < \varepsilon\).
This provides a precise way of defining what it means for \(f(x)\) to get arbitrarily close to \(L\) as \(x\) gets arbitrarily close to \(c\). In the given exercise, the \(\varepsilon\) is used to dictate how close \(2x + 8\) must be to zero, establishing a relationship with \(|x+4|<\frac{\varepsilon}{2}\) that maintains these conditions.
Mastering Inequalities in Calculus
Inequalities in calculus can often guide us to understand and solve more complex problems. They express a range of possible values rather than one precise value, providing flexibility in approaching solutions.
When we consider the inequality \(|x+4|<\frac{\varepsilon}{2}\), it gives a boundary on the magnitude of \(x+4\). This boundary simplifies problem scenarios by narrowing down options, helping us assert that a statement might be universally true as long as a certain condition is met.
In calculus, establishing these inequalities and consistently transforming them as shown—where \(|2(x+4)| = 2 |x+4| < \varepsilon\)—proves to be highly valuable in proving implications and understanding limits, continuity, and other key calculus concepts.
When we consider the inequality \(|x+4|<\frac{\varepsilon}{2}\), it gives a boundary on the magnitude of \(x+4\). This boundary simplifies problem scenarios by narrowing down options, helping us assert that a statement might be universally true as long as a certain condition is met.
In calculus, establishing these inequalities and consistently transforming them as shown—where \(|2(x+4)| = 2 |x+4| < \varepsilon\)—proves to be highly valuable in proving implications and understanding limits, continuity, and other key calculus concepts.
Understanding Function Transformations
Function transformations involve shifting, stretching, or compressing a graph of a function, and they are often key in solving and visualizing calculus problems.
When we transform functions, like expressing \(2x+8\) as \(2(x+4)\), we perform a crucial operation that maintains equivalency while enabling more direct analysis.
Transformations can simplify expressions and link the condition directly to the outcome, as shown by rewriting and simplifying expressions in terms of constants or coefficients. In the example, the transformation simplifies the inequality math and shows how the conditions set by the absolute value \(|x+4|\) then transition directly into the desired format \(|2x+8|<\varepsilon\), providing clarity and efficiency in finding solutions.
When we transform functions, like expressing \(2x+8\) as \(2(x+4)\), we perform a crucial operation that maintains equivalency while enabling more direct analysis.
Transformations can simplify expressions and link the condition directly to the outcome, as shown by rewriting and simplifying expressions in terms of constants or coefficients. In the example, the transformation simplifies the inequality math and shows how the conditions set by the absolute value \(|x+4|\) then transition directly into the desired format \(|2x+8|<\varepsilon\), providing clarity and efficiency in finding solutions.
