Let’s solve each part step-by-step. **Since you asked for sketches, I'll describe how you would draw each graph, and explain the reasoning behind each type of discontinuity.** --- ### **a. \( f(x) \) is continuous for all \( x \neq 2 \) and has a removable discontinuity at \( x = 2 \).** #### **Explanation:** - A removable discontinuity occurs when the limit exists at a point, but the function is either not defined there, or defined to a different value. - So, \( f(x) \) could be any continuous function except at \( x=2 \). #### **Example function:** \[ f(x) = \frac{x^2 - 4}{x-2} \] - For \( x \neq 2 \), \( f(x) = x+2 \). - At \( x=2 \), the function is undefined (hole in the graph). #### **Graph Sketch:** - Draw the line \( y=x+2 \), but put a **hole** at \( x = 2 \). --- ### **b. The domain of \( f \) is \( [0, 4) \cup (4, \infty) \), \( f \) is continuous from the right at \( x = 0 \), continuous on \( 0 < x < 4 \), and continuous for \( x > 4 \).** #### **Explanation:** - The function is not defined at \( x = 4 \). - Right-continuous at \( x = 0 \) means \( \lim_{x \to 0^+} f(x) = f(0) \). #### **Example function:** \[ f(x) = \begin{cases} x^2 & 0 \leq x < 4 \\ x + 2 & x > 4 \end{cases} \] - At \( x = 0 \), \( f(0) = 0 \), and the right limit as \( x \to 0^+ \) is 0. - Not defined at \( x = 4 \). #### **Graph Sketch:** - Draw \( y = x^2 \) from \( x = 0 \) (solid dot) to just before \( x = 4 \) (open dot). - Draw \( y = x+2 \) for \( x > 4 \). - No point at \( x = 4 \). --- ### **c. \( f \) is continuous for all \( x \neq 1 \), has a removable discontinuity at \( x = 1 \), and an infinite discontinuity at \( x = 3 \).** #### **Explanation:** - Removable at \( x = 1 \): hole in the graph. - Infinite at \( x = 3 \): vertical asymptote. #### **Example function:** \[ f(x) = \frac{(x-1)}{(x-3)} \] - At \( x = 1 \), \( f(1) = 0 \), but you can redefine \( f(1) \) to be something else to make it removable. - At \( x = 3 \), denominator is zero, so vertical asymptote. #### **Graph Sketch:** - Draw the hyperbola \( y = \frac{x-1}{x-3} \). - Place a **hole** at \( x = 1 \). - Draw a **vertical asymptote** at \( x = 3 \). --- ### **d. \( f(2) = -3 \), \( f \) is continuous for all \( x \neq 1, 3 \), has a jump discontinuity at \( x = 3 \), and an infinite discontinuity at \( x = 1 \).** #### **Explanation:** - At \( x = 1 \): vertical asymptote. - At \( x = 3 \): jump in the graph (left and right limits are different). - At \( x = 2 \): specific value \( f(2) = -3 \). - Elsewhere: continuous. #### **Example function:** \[ f(x) = \begin{cases} \frac{x-2}{x-1} & x < 3 \\ \frac{x+1}{x-1} & x > 3 \end{cases} \] - At \( x = 1 \): denominator zero, infinite discontinuity. - At \( x = 3 \): left and right limits are different (jump). - At \( x = 2 \): \( f(2) = 0 \), so redefine to \( f(2) = -3 \). #### **Graph Sketch:** - Draw \( y = \frac{x-2}{x-1} \) for \( x < 3 \), with a **vertical asymptote** at \( x = 1 \). - Draw \( y = \frac{x+1}{x-1} \) for \( x > 3 \), also with a **vertical asymptote** at \( x = 1 \). - At \( x = 3 \): plot open circles at the left and right limit values (jump). - At \( x = 2 \): plot a solid point at \( (2, -3) \), and open circle at the value from the formula. --- **Let me know if you'd like detailed sketches for each!**