Questions: 11.9 HW - Continuity and Differentiability Question 8, 11.9.27 Part 2 of 2 (a) Graph the given function. (b) Find all values of x where the function is discontinuous. g(x) = 13 if x < -1 x^2 + 4 if -1 ≤ x ≤ 3 13 if x > 3 (a) Choose the correct graph below. A. B. (b) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is discontinuous at x = . (Use a comma to separate answers as needed.) B. The function is continuous for all values of x.

Solution Steps

The given function \( g(x) \) is defined as: \[ g(x) = \begin{cases} 13 & \text{if } x < -1 \\ x^2 + 4 & \text{if } -1 \leq x \leq 3 \\ 13 & \text{if } x > 3 \end{cases} \]

Potential points of discontinuity occur at the boundaries of the piecewise function, which are \( x = -1 \) and \( x = 3 \).

• For \( x < -1 \), \( g(x) = 13 \).
• For \( x = -1 \), \( g(x) = (-1)^2 + 4 = 5 \).

Since \( g(x) \) jumps from 13 to 5 at \( x = -1 \), the function is discontinuous at \( x = -1 \).

• For \( -1 \leq x \leq 3 \), \( g(x) = x^2 + 4 \).
• For \( x = 3 \), \( g(x) = 3^2 + 4 = 13 \).
• For \( x > 3 \), \( g(x) = 13 \).

Since \( g(x) is continuous at \( x = 3 \) (no jump in value), the function is continuous at \( x = 3 \).

The correct graph should reflect the discontinuity at \( x = -1 \) and continuity at \( x = 3 \). Graph B shows this behavior.

Final Answer