For the given function​ f(x) and values of​ L, 

cc​, 

and 

epsilonεgreater than>0 

find the largest open interval about 

cc 

on which the inequality 

StartAbsoluteValue f left parenthesis x right parenthesis minus Upper L EndAbsoluteValue less than epsilonf(x)−L<ε 

holds. Then determine the largest value for 

deltaδgreater than>0 

such that for all x satisfying 

0less than<StartAbsoluteValue x minus c EndAbsoluteValuex−cless than<deltaδ​, 

the inequality 

StartAbsoluteValue f left parenthesis x right parenthesis minus Upper L EndAbsoluteValue less than epsilonf(x)−L<ε 

holds.

​f(x)equals=StartRoot 7 x plus 86 EndRoot7x+86​, 
  
Lequals=​11, 
  
ccequals=​5, 
  
epsilonεequals=0.070.07

Question content area bottom

Part 1

The largest open interval about 

cc 

on which the inequality 

StartAbsoluteValue f left parenthesis x right parenthesis minus Upper L EndAbsoluteValue less than epsilonf(x)−L<ε 

holds is left parenthesis nothing comma nothing right parenthesis .enter your response here,enter your response here.

​(Round to four decimal places as​ needed.)

To solve this problem, we need to follow a logical process and break it down into steps.

### Given:
- \( f(x) = \sqrt{7x + 86} \)
- \( L = 11 \)
- \( c = 5 \)
- \( \epsilon = 0.07 \)

### Step 1: Determine \( f(c) \)
We start by calculating \( f(c) \) where \( c = 5 \).

\[
f(5) = \sqrt{7(5) + 86} = \sqrt{35 + 86} = \sqrt{121} = 11
\]
Thus, \( f(5) = 11 \), which matches the given \( L \).

### Step 2: Set up the inequality
We are given the inequality:

\[
| f(x) - L | < \epsilon
\]

Substitute the given values of \( f(x) \), \( L \), and \( \epsilon \):

\[
| \sqrt{7x + 86} - 11 | < 0.07
\]

### Step 3: Solve the inequality
We now need to solve the inequality \( | \sqrt{7x + 86} - 11 | < 0.07 \).

This can be split into two inequalities:

\[
-0.07 < \sqrt{7x + 86} - 11 < 0.07
\]

Adding 11 to all parts:

\[
10.93 < \sqrt{7x + 86} < 11.07
\]

### Step 4: Square both sides
To eliminate the square root, square all parts of the inequality:

\[
(10.93)^2 < 7x + 86 < (11.07)^2
\]

Calculating the squares:

\[
119.0649 < 7x + 86 < 122.5449
\]

### Step 5: Isolate \( x \)
Now subtract 86 from all parts:

\[
33.0649 < 7x < 36.5449
\]

Finally, divide by 7:

\[
4.7236 < x < 5.2207
\]

### Step 6: Largest open interval
The largest open interval about \( c = 5 \) on which the inequality holds is:

\[
(4.7236, 5.2207)
\]

### Step 7: Find \( \delta \)
To determine \( \delta \), we need to find the distance from \( c = 5 \) to the endpoints of the interval:

\[
\delta = \min(5 - 4.7236, 5.2207 - 5)
\]

Calculating both:

\[
5 - 4.7236 = 0.2764
\]
\[
5.2207 - 5 = 0.2207
\]

Thus, \( \delta = 0.2207 \).

### Final Answers:
- The largest open interval is \( (4.7236, 5.2207) \).
- The largest \( \delta \) is \( 0.2207 \).

Would you like more details on any part of the process?

---

Here are 5 related questions for further practice:
1. How does changing \( \epsilon \) affect the value of \( \delta \)?
2. What happens if \( f(x) \) is a different function, such as \( f(x) = 2x + 5 \)?
3. How can you interpret \( \delta \) geometrically on a graph?
4. How does this problem relate to the formal definition of limits?
5. What if \( f(x) \) involved a higher-order polynomial?

**Tip:** Always carefully isolate \( x \) when solving inequalities to ensure the correct interval.

For the given function f(x) and values of L, c, and ε > 0, find the largest open interval about c on which the inequality |f(x) - L| < ε holds. Then determine the largest value for δ > 0 such that for all x satisfying 0 < |x - c| < δ, the inequality |f(x) - L| < ε holds. f(x) = √(7x + 86), L = 11, c = 5, ε = 0.07.

Explore a detailed solution to find the largest open interval for the function √(7x + 86) where |f(x) - L| < ε and determine the largest delta (δ) value. This calculus problem involves concepts of limits, continuity, and inequality solving, suitable for students in Grades 11-12.

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