Questions: (a) Graph the function (b) find all values of (x) where the function is discontinuous, and (c) find the limit from the left and the right at any values of (x) where the function is discontinuous. [ g(x)=leftbeginarrayll -2 x<-3 x^2-3 x<1 -2 x>1 endarrayright. ] (a) Choose the correct graph of the function. A. B. C. (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 piecewise function is: \[ g(x) = \begin{cases} -2 & \text{if } x < -3 \\ x^2 - 3 & \text{if } -3 \le x \le 1 \\ -2 & \text{if } x > 1 \end{cases} \] We analyze the function at the points where the pieces change, which are \(x = -3\) and \(x = 1\).

For \(x = -3\): \[g(-3) = (-3)^2 - 3 = 9 - 3 = 6\] \[\lim_{x \to -3^-} g(x) = -2\] \[\lim_{x \to -3^+} g(x) = (-3)^2 - 3 = 6\] Since the left-hand limit and the right-hand limit are not equal, the function is discontinuous at \(x = -3\).

For \(x = 1\): \[g(1) = (1)^2 - 3 = 1 - 3 = -2\] \[\lim_{x \to 1^-} g(x) = (1)^2 - 3 = -2\] \[\lim_{x \to 1^+} g(x) = -2\] Since the left-hand limit, the right-hand limit, and the function value at \(x = 1\) are all equal to -2, the function is continuous at \(x = 1\).

The correct graph is B. Graph A shows a discontinuity at \(x=1\) but not \(x=-3\). Graph C shows a discontinuity at \(x=-3\) but appears to be \(x^2\) rather than \(x^2-3\). Graph B shows a discontinuity at \(x=-3\), and the portion for \(-3 \le x \le 1\) resembles the graph of \(x^2 - 3\). The constant \(y=-2\) is represented correctly on either side.

The function is discontinuous at \(x = -3\). At \(x = -3\): Left-hand limit: \(\lim_{x \to -3^-} g(x) = -2\) Right-hand limit: \(\lim_{x \to -3^+} g(x) = 6\)

Final Answer