Question

recall that a function f is continuous at a number a if the following holds.
$$\lim_{x \to a} f(x) = f(a)$$
we are given the following.

$$f(x) = \
LATEXBLOCK0
$$

we note that for all values of the constant c the function f is continuous on both $(-\infty, 6)$ and $(6, \infty)$. the only case that we need to consider is when $x = 6$. for the function f to be continuous, we need to ensure that there is not a discontinuity at $x = 6$. to do so, we must determine the value of c such that the following holds.
$$\lim_{x \to 6^-} f(x) = \lim_{x \to 6^+} f(x)$$
find the following limits.
$$\lim_{x \to 6^-} f(x) = \lim_{x \to 6^-} (cx^2 + 5x) = \square$$
$$\lim_{x \to 6^+} f(x) = \lim_{x \to 6^+} (x^3 - cx) = \square$$

Explanation:

Step1: Calculate left - hand limit

To find \(\lim_{x
ightarrow6^{-}}f(x)=\lim_{x
ightarrow6^{-}}(cx^{2}+5x)\), we substitute \(x = 6\) into the function \(cx^{2}+5x\) (since the function is a polynomial and polynomials are continuous everywhere, so the left - hand limit as \(x
ightarrow6\) is the value of the function at \(x = 6\)).
\(\lim_{x
ightarrow6^{-}}(cx^{2}+5x)=c\times6^{2}+5\times6=c\times36 + 30=36c + 30\)

Step2: Calculate right - hand limit

To find \(\lim_{x
ightarrow6^{+}}f(x)=\lim_{x
ightarrow6^{+}}(x^{3}-cx)\), we substitute \(x = 6\) into the function \(x^{3}-cx\) (since the function is a polynomial and polynomials are continuous everywhere, so the right - hand limit as \(x
ightarrow6\) is the value of the function at \(x = 6\)).
\(\lim_{x
ightarrow6^{+}}(x^{3}-cx)=6^{3}-c\times6=216-6c\)

Answer:

For \(\lim_{x
ightarrow6^{-}}f(x)\), the answer is \(36c + 30\); for \(\lim_{x
ightarrow6^{+}}f(x)\), the answer is \(216-6c\)