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