Example6.4.3(a)Provethatf:R→Rdefinedby
22x+1[3(5)22(5)+1]
35
3x+5+2<3(|x|+5)+2<3(6+5)+2=35.
enableustousetheconditionthat|x5|<δ.So,continuing
Oncewehavefactoredoutoneδfromf(x)−f(5)|,we
wouldliketoestimateanyfactorassociatedwiththisδby
|f(x)f(5)|=|3x
≤3δx+5|+2δ=(3x+5+2)δ.
Observethatwedidnotusethenumericalvalueoff(5)=
Recallthatwealreadymadetherestrictionthatδ<1.
2−5
.
Note:Thetrickinfindingaδ>0liesintheuseofthegiven
|f(x)−f(5)≤3|x−5·x+5+2|x−5
Hence,
=|3(x
Solution:Let>0begiven.Wewanttofindaδ=δ()>
2−2x+1iscontinuousatx=5.
)−2(x−5).
So,given>0,wechooseδ=min{1,
of|f(x)−f(5).Now,
dothis.Now,fromx−5|<δwehave
wehavethat
afactorofx5inourestimateoff(x)f(5)|.Thiswill
conditionthatx5<δ,andinthemethodofevaluation
2

3(52
0suchthatifx5|<δthenf(x)f(5)|<.
f(x)=3x
Herewecanassumeδ<1sothat|x|<6.Withthis,
x=x5+5|≤x5|+5<δ+5.
35}.Then,
aninteger.Wealwaysusethegivencondition|x5<δto

)−2(5)+1=66.Thisisbecausewewouldliketoget
|f(x)f(5)≤(3|x+5+2)δ<35δ=ifδ=
f(x)−f(5)<wheneverx−5|<δ,

It seems you're referencing a proof related to the continuity of a function \( f(x) = 3x + 5 \) at \( x = 5 \), using the epsilon-delta definition of continuity. Let me clarify the key steps involved, and how to correctly write out the proof.

### Problem:
We want to show that the function \( f(x) = 3x + 5 \) is continuous at \( x = 5 \) using the epsilon-delta definition of continuity.

### Solution Outline:

To prove that \( f(x) = 3x + 5 \) is continuous at \( x = 5 \), we use the epsilon-delta definition of continuity:
\[
\lim_{x \to 5} f(x) = f(5).
\]
This is equivalent to showing that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( |x - 5| < \delta \), then \( |f(x) - f(5)| < \epsilon \).

### Step-by-Step Solution:

1. **Define the function**:
   The function is given by \( f(x) = 3x + 5 \), and we want to show that it is continuous at \( x = 5 \).

2. **Compute \( f(5) \)**:
   \[
   f(5) = 3(5) + 5 = 15 + 5 = 20.
   \]

3. **Consider the difference \( |f(x) - f(5)| \)**:
   We need to analyze the expression:
   \[
   |f(x) - f(5)| = |(3x + 5) - 20| = |3x - 15| = 3|x - 5|.
   \]
   So, we have:
   \[
   |f(x) - f(5)| = 3|x - 5|.
   \]

4. **Set up the condition**:
   We want \( |f(x) - f(5)| < \epsilon \). From the previous step, we know:
   \[
   3|x - 5| < \epsilon.
   \]
   Dividing both sides by 3:
   \[
   |x - 5| < \frac{\epsilon}{3}.
   \]

5. **Choose \( \delta \)**:
   To satisfy the condition \( |x - 5| < \delta \) whenever \( |f(x) - f(5)| < \epsilon \), we set:
   \[
   \delta = \frac{\epsilon}{3}.
   \]
   This ensures that if \( |x - 5| < \delta \), then \( |f(x) - f(5)| < \epsilon \).

6. **Conclusion**:
   We have shown that for every \( \epsilon > 0 \), there exists \( \delta = \frac{\epsilon}{3} \) such that if \( |x - 5| < \delta \), then \( |f(x) - f(5)| < \epsilon \). Therefore, \( f(x) = 3x + 5 \) is continuous at \( x = 5 \).

### Summary:

The function \( f(x) = 3x + 5 \) is continuous at \( x = 5 \), and the epsilon-delta argument shows that for any \( \epsilon > 0 \), we can find a corresponding \( \delta = \frac{\epsilon}{3} \) such that \( |f(x) - f(5)| < \epsilon \) when \( |x - 5| < \delta \).

---

Would you like any further clarification on the steps, or any specific questions about the method used? Here are a few related questions to explore more:

1. How does the epsilon-delta definition of continuity work in other types of functions?
2. What is the geometric interpretation of the epsilon-delta proof for continuity?
3. How does the proof change for more complex functions (e.g., quadratic or piecewise functions)?
4. What happens if we try to prove continuity at a point where the function is not continuous?
5. Can you prove the continuity of \( f(x) = x^2 \) at \( x = 3 \) using a similar method?

**Tip**: For functions that are simple polynomials (like \( f(x) = 3x + 5 \)), the proof is often straightforward. But for more complicated functions, always start by simplifying the expression \( |f(x) - f(c)| \) before attempting to find \( \delta \).

Example 6.4.3(a) Prove that f: R → R defined by f(x) = 3x + 5 is continuous at x = 5 using the epsilon-delta definition of continuity.

Understand how to prove the continuity of the linear function f(x) = 3x + 5 at x = 5 using the epsilon-delta definition of continuity. Includes step-by-step reasoning, formula derivation, and key mathematical concepts. Ideal for students in Grades 11-12.

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