To determine the value of $c$ that makes the function $f(x)$ continuous on $(-\infty, \infty)$, we need to ensure that the function is continuous at $x = 5$ since this is the point where the definition of $f(x)$ changes.

Given:

• For $x < 5$: $f(x) = cx^2 + 6x$
• For $x \geq 5$: $f(x) = x^3 - cx$

Continuity condition:

For $f(x)$ to be continuous at $x = 5$, the left-hand limit as $x \to 5^-$ must equal the right-hand limit as $x \to 5^+$, and both must equal $f(5)$.

1. Left-hand limit (as $x \to 5^-$):

$\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (cx^2 + 6x) = c(5^2) + 6(5) = 25c + 30$

2. Right-hand limit (as $x \to 5^+$):

$\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x^3 - cx) = 5^3 - c(5) = 125 - 5c$

For continuity at $x = 5$, these two limits must be equal:

$25c + 30 = 125 - 5c$

Solve for $c$:

$25c + 5c = 125 - 30$ $30c = 95$ $c = \frac{95}{30} = \frac{19}{6}$

Thus, the value of $c$ that makes $f(x)$ continuous on $(-\infty, \infty)$ is $\boxed{\frac{19}{6}}$.

Would you like more details or have any questions?

Here are five related questions:

1. How do you check for differentiability at $x = 5$ for this function?
2. What happens if $c$ takes a different value than $\frac{19}{6}$?
3. Can you graph the function to visualize continuity at $x = 5$?
4. What if the function had different expressions for $f(x)$ around $x = 5$?
5. How do you compute the derivative of piecewise functions?

Tip: Always check both left-hand and right-hand limits when testing for continuity at a specific point in piecewise functions.