To find the value of $c$ that makes the function continuous, we need to ensure that the two pieces of the function meet at $x = 3$. The two parts of the function are:

• $f(x) = x^2 - c$ for $x < 3$
• $f(x) = 6x + 5c$ for $x \geq 3$

For the function to be continuous at $x = 3$, the left-hand limit as $x \to 3^-$ must equal the right-hand limit as $x \to 3^+$, and both must equal the value of the function at $x = 3$.

We will set up the equation:

$\lim_{x \to 3^-} (x^2 - c) = \lim_{x \to 3^+} (6x + 5c)$

1. The left-hand limit is: $\lim_{x \to 3^-} (x^2 - c) = 3^2 - c = 9 - c$

2. The right-hand limit is: $\lim_{x \to 3^+} (6x + 5c) = 6(3) + 5c = 18 + 5c$

Now, set these two expressions equal to each other to ensure continuity:

$9 - c = 18 + 5c$

Solve for $c$:

$9 - 18 = 5c + c$ $-9 = 6c$ $c = -\frac{3}{2}$

Thus, the value of $c$ that makes the function continuous is $c = -\frac{3}{2}$.

Would you like further details on any part of the solution, or do you have any questions?

Here are 5 related questions:

1. How do we determine if a function is continuous at a point?
2. What happens if the limits on both sides of a function do not match?
3. How do you handle piecewise functions in limit problems?
4. How would you solve for a constant in a piecewise function to ensure differentiability?
5. What is the difference between continuity and differentiability?

Tip: When solving continuity problems for piecewise functions, always ensure the limits from both sides match, not just the function value!