Questions: Consider the function g defined by g(x)=c/(x-17) for x<10, C+3x for x=10, 2C-4 for x>10. The symbol C represents the same real number in each of the piecewise formulas. Find the value for C that makes the function continuous on (-∞, 10]. Make sure that your reasoning is clear. Is it possible to find a value for C that makes the function continuous over (-∞, ∞) ? Explain.

Solution Steps

To make the function continuous on $(- \infty, 10]$, we need to ensure that the left-hand limit as $x$ approaches 10 from the left is equal to the value of the function at $x = 10$. This means solving the equation $\lim_{x \to 10^-} \frac{c}{x-17} = C + 3 \times 10$.

To ensure the function $g(x)$ is continuous at $x = 10$, the left-hand limit as $x$ approaches 10 from the left must equal the value of the function at $x = 10$.

The function is defined as:

• $g(x) = \frac{c}{x-17}$ for $x < 10$
• $g(x) = C + 3x$ for $x = 10$

The left-hand limit as $x \to 10^-$ is: $$\lim_{x \to 10^-} g(x) = \lim_{x \to 10^-} \frac{c}{x-17}$$

Substituting $x = 10$ into the expression: $$\lim_{x \to 10^-} \frac{c}{x-17} = \frac{c}{10-17} = \frac{c}{-7}$$

The value of the function at $x = 10$ is: $$g(10) = C + 3 \times 10 = C + 30$$

For continuity at $x = 10$: $$\frac{c}{-7} = C + 30$$

Rearrange the equation for $C$: $$C = \frac{c}{-7} - 30$$

For the function to be continuous over $(-\infty, \infty)$, the right-hand limit as $x \to 10^+$ must also equal the value of the function at $x = 10$.

The function is defined as:

• $g(x) = 2C - 4$ for $x > 10$

The right-hand limit as $x \to 10^+$ is: $$\lim_{x \to 10^+} g(x) = 2C - 4$$

For continuity at $x = 10$: $$2C - 4 = C + 30$$

Rearrange the equation: $$2C - C = 30 + 4$$ $$C = 34$$

Final Answer