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?

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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.