Final answer:
The piecewise function has 1 discontinuity, which occurs at x = 5, since there is a mismatch in the function values from either side of this point.
Step-by-step explanation:
To determine how many discontinuities the piecewise function has, we need to examine each interval transition point to see if the function values match up from either side of the point:
- At x = -3, the transition from 3ex-3 + 1 to (2/3)x2 - 1 needs to be continuous.
- At x = 3, the transition from (2/3)x2 - 1 to -7/2x + 29/2 needs to be continuous.
- At x = 5, the transition from -7/2x + 29/2 to log(2x-4) needs to be continuous.
We need to check each interval's endpoints by substituting the boundary values into the corresponding functions:
- When x = -3, we get 3e0 + 1 = 4 from the first function and (2/3)(-3)2 - 1 = 4 from the second function, so they match, and there is no discontinuity at x = -3.
- For x = 3, the second function gives (2/3)(3)2 - 1 = 5 and the third function gives -7/2(3) + 29/2 = 5. There's no discontinuity at x = 3.
- When x = 5, the third function yields -7/2(5) + 29/2 = 2, and the fourth function log(2(5)-4) = log(6) is not equal to 2, which means there is a discontinuity at x = 5.
Therefore, the piecewise function has 1 discontinuity.