Given f f g are Continuous at a ERP So, for € 70 ISO such that Thus 창 / & for g, J 8₂ 70 ) f(x)-f(a)|
Given f f g are Continuous at a ERP So, for € 70 ISO such that Thus 창 / & for g, J 8₂ 70 ) f(x)-f(a)|
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Write thoroughly explaining in detailed WORDS ONLY the solution in these photos- (Words explaining the solution to a non-math major)
![**Continuity of Functions**
**Given:**
Functions \( f \) and \( g \) are continuous at \( a \in \mathbb{R}^p \).
**For \( f \):**
For every \( \varepsilon > 0 \), there exists \( \delta_1 > 0 \) such that
\[ |f(x) - f(a)| < \varepsilon \quad \forall \|x - a\| < \delta_1 \]
**For \( g \):**
There exists \( \delta_2 > 0 \) such that
\[ |g(x) - g(a)| < \varepsilon \quad \forall \|x - a\| < \delta_2 \]
Thus, if we choose \( \delta = \min\{\delta_1, \delta_2\} \), then
\[ |f(x) - f(a)| < \varepsilon \quad \forall \|x - a\| < \delta_1, \delta < \delta \]
\[ |g(x) - g(a)| < \varepsilon \quad \forall \|x - a\| < \delta_2, \delta < \delta \]
**Conclusion:**
Hence, for our chosen \( \varepsilon > 0 \), we get \( \delta > 0 \) such that
\[ |f(x) - f(a)| < \varepsilon \]
and
\[ |g(x) - g(a)| < \varepsilon \]
whenever \(\|x - a\| < \delta\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff5203d42-0da6-4b1b-a1a9-5b37798d2cd6%2Fb24b37be-a382-458b-a544-c63d6073873b%2Fxxluq4i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Continuity of Functions**
**Given:**
Functions \( f \) and \( g \) are continuous at \( a \in \mathbb{R}^p \).
**For \( f \):**
For every \( \varepsilon > 0 \), there exists \( \delta_1 > 0 \) such that
\[ |f(x) - f(a)| < \varepsilon \quad \forall \|x - a\| < \delta_1 \]
**For \( g \):**
There exists \( \delta_2 > 0 \) such that
\[ |g(x) - g(a)| < \varepsilon \quad \forall \|x - a\| < \delta_2 \]
Thus, if we choose \( \delta = \min\{\delta_1, \delta_2\} \), then
\[ |f(x) - f(a)| < \varepsilon \quad \forall \|x - a\| < \delta_1, \delta < \delta \]
\[ |g(x) - g(a)| < \varepsilon \quad \forall \|x - a\| < \delta_2, \delta < \delta \]
**Conclusion:**
Hence, for our chosen \( \varepsilon > 0 \), we get \( \delta > 0 \) such that
\[ |f(x) - f(a)| < \varepsilon \]
and
\[ |g(x) - g(a)| < \varepsilon \]
whenever \(\|x - a\| < \delta\).
![### Continuity of the Product of Functions
To show that \( f \cdot g \) is continuous at \( x = a \), we begin with:
\[ | (f \cdot g)(x) - (f \cdot g)(a) | \]
This is expressed as:
\[ = | f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a) | \]
Using the triangle inequality:
\[ \leq | f(x)g(x) - f(x)g(a) | + | f(x)g(a) - f(a)g(a) | \]
Further simplifying:
\[ \leq | f(x) || g(x) - g(a) | + | g(a) || f(x) - f(a) | \]
Now, since \( f \) is continuous, it is bounded on closed balls. Therefore, if we take \( \tilde{\delta} < \delta \), then:
\[ | f(x) | \leq M \quad \text{for} \quad |x - a| \leq \tilde{\delta} \quad \text{for some} \quad M > 0 \]
Using this, the inequality becomes:
\[ |(f \cdot g)(x) - (f \cdot g)(a)| \leq M \epsilon + | g(a) | \epsilon = \epsilon (M + g(a)) \]
whenever \( |x - a| \leq \tilde{\delta} \).
Clearly, if we choose \( \epsilon_1 > 0 \) and let \( \epsilon = \frac{\epsilon_1}{M + g(a)} \), then we get some \( \delta' > 0 \) [in fact we can take \(\delta' = \tilde{\delta}\)] such that:
\[ |(f \cdot g)(x) - (f \cdot g)(a)| < \epsilon_1, \quad \text{whenever} \quad |x - a| < \delta' \]
Thus, \( f \cdot g \) is continuous at \( x = a \).
**(Proved)**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff5203d42-0da6-4b1b-a1a9-5b37798d2cd6%2Fb24b37be-a382-458b-a544-c63d6073873b%2Fdd6rko_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Continuity of the Product of Functions
To show that \( f \cdot g \) is continuous at \( x = a \), we begin with:
\[ | (f \cdot g)(x) - (f \cdot g)(a) | \]
This is expressed as:
\[ = | f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a) | \]
Using the triangle inequality:
\[ \leq | f(x)g(x) - f(x)g(a) | + | f(x)g(a) - f(a)g(a) | \]
Further simplifying:
\[ \leq | f(x) || g(x) - g(a) | + | g(a) || f(x) - f(a) | \]
Now, since \( f \) is continuous, it is bounded on closed balls. Therefore, if we take \( \tilde{\delta} < \delta \), then:
\[ | f(x) | \leq M \quad \text{for} \quad |x - a| \leq \tilde{\delta} \quad \text{for some} \quad M > 0 \]
Using this, the inequality becomes:
\[ |(f \cdot g)(x) - (f \cdot g)(a)| \leq M \epsilon + | g(a) | \epsilon = \epsilon (M + g(a)) \]
whenever \( |x - a| \leq \tilde{\delta} \).
Clearly, if we choose \( \epsilon_1 > 0 \) and let \( \epsilon = \frac{\epsilon_1}{M + g(a)} \), then we get some \( \delta' > 0 \) [in fact we can take \(\delta' = \tilde{\delta}\)] such that:
\[ |(f \cdot g)(x) - (f \cdot g)(a)| < \epsilon_1, \quad \text{whenever} \quad |x - a| < \delta' \]
Thus, \( f \cdot g \) is continuous at \( x = a \).
**(Proved)**
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