Question

we have determined that
$$\lim_{x \to 6^-} f(x) = 36c + 30 \text{ and } \lim_{x \to 6^+} f(x) = 216 - 6c.$$
we can use these to solve for $c$.
$$\lim_{x \to 6^-} f(x) = \lim_{x \to 6^+} f(x)$$
$$36c + 30 = \square - 6c$$
$$c = \square$$
therefore, for what value of the constant $c$ is the function continuous on $(-\infty, \infty)$?
$$c = \square$$
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Explanation:

Step1: Substitute the right - hand limit

We know that \(\lim_{x
ightarrow6^{-}}f(x)=36c + 30\) and \(\lim_{x
ightarrow6^{+}}f(x)=216-6c\). When we set \(\lim_{x
ightarrow6^{-}}f(x)=\lim_{x
ightarrow6^{+}}f(x)\), we substitute the expression for \(\lim_{x
ightarrow6^{+}}f(x)\) into the equation. So \(36c + 30=216-6c\).

Step2: Solve for \(c\)

First, we add \(6c\) to both sides of the equation \(36c + 30=216-6c\) to get \(36c+6c + 30=216-6c + 6c\), which simplifies to \(42c+30 = 216\). Then we subtract 30 from both sides: \(42c+30 - 30=216 - 30\), so \(42c=186\). Next, we divide both sides by 42: \(c=\frac{186}{42}=\frac{31}{7}\approx4.4286\)? Wait, no, let's recalculate. Wait, \(216 - 30 = 186\)? Wait, \(216-30 = 186\), and \(42c = 186\), then \(c=\frac{186}{42}=\frac{31}{7}\)? Wait, no, \(186\div6 = 31\), \(42\div6 = 7\), yes. Wait, but let's check again. \(36c+6c=42c\), \(216 - 30 = 186\), so \(c=\frac{186}{42}=\frac{31}{7}\approx4.4286\)? Wait, no, maybe I made a mistake. Wait, \(216-30 = 186\)? Wait, \(216 - 30=186\), and \(42c = 186\), so \(c=\frac{186}{42}=\frac{31}{7}\)? Wait, no, \(186\div6 = 31\), \(42\div6 = 7\), so \(c = \frac{31}{7}\)? Wait, but let's do the arithmetic again. \(36c+30=216 - 6c\). Add \(6c\) to both sides: \(42c+30 = 216\). Subtract 30: \(42c=186\). Then \(c=\frac{186}{42}=\frac{31}{7}\approx4.4286\)? Wait, no, \(186\div42 = 4.42857\cdots\), but maybe I made a mistake in the problem. Wait, the right - hand limit is \(216-6c\), so the first blank is 216. Then solving \(36c + 30=216-6c\):

\(36c+6c=216 - 30\)

\(42c=186\)

\(c=\frac{186}{42}=\frac{31}{7}\approx4.4286\)? Wait, no, \(186\div6 = 31\), \(42\div6 = 7\), so \(c=\frac{31}{7}\). Wait, but let's check with another approach. \(36c+6c=42c\), \(216 - 30 = 186\), so \(c=\frac{186}{42}=\frac{31}{7}\).

Answer:

First blank: \(216\); \(c=\frac{31}{7}\) (or approximately \(4.43\) if decimal is preferred, but the exact value is \(\frac{31}{7}\))