Number 85. Which of the following statements is true? So we have a peacewise function. And let's go through each answer choice. Looking at answer choice A. F ofx is continuous at x= 5. Is that true or false? Well, if we take the limit as x approaches 5 from the left side. So we need to use this part of the peace-wise function. That's when x is less than 5. We'll get 5^ 2 - 4 * 5 - 12. 5^2 is 25. 4 * 5 is 20. 25 - 20 is 5. 5 - 12 is 7. So as x approaches 5 from the left we get a value of 7. Now notice that for this peacewise function x also equals 5. What that means is that f of five will also be 7 because we'll have to plug it into this to get that answer. So f of five is defined. Now let's check the limit as x approaches 5 on the right side. So to the right of five where x is greater than 5 we need to use this one 3x - 23. So this is going to be 3 * 5 - 23 3 * 5 is 15. 15 - 23 is8. So notice that the leftsided limit and the right sided limit they do not match. Therefore the limit as x approaches 5 in general of f ofx does not exist because there's a mismatch. What we have here is a jump discontinuity. By the way, this should be -7 instead of8. I don't know why I have that because this is 25 - 20. That's 5. 5 - 12 is -7. So, because the leftsided limit and the right sided limit doesn't match, we have a jump discontinuity. So, A is a false statement. It is not continuous at X= 5. Now, what about B? f ofx has a removable discontinuity at x= 5. If it was removable, these two would be the same, but this would be different. The leftsided limit and the right side limit would match. The limit would exist, but f of a just wouldn't equal that limit. That would be a whole or removable discontinuity. But because the leftsided and the right sided limit do not match. What we have here is a jump is discontinuity. So when x is five on the left the y value will be at -7. And when x is five on the right or greater than five the y value is8. As you could see there's a jump discontinuity here. But at -7 uh we have a closed circle because it includes five and this will be an open circle. But because of the disconnect is the jump discontinuity and jump discontinuities are non removable discontinuities. So that's why B is false. What we have is a non removable discontinuity, not a removable one. Now what about answer choice C? F of 5 is equal to 8. But we know that's false. F of 5 is -7. Now what about D? f ofx is continuous at x= 9. So let's find out. Let's take the limit as x approaches 9 from the left. So less than 9, we would have to use this one. 3x - 23. So that's going to be 3 * 9 - 23. 3 * 9 is 27. 27 - 23 is 4. Now notice that this includes 9. What that means is that f of 9 will also equal 4. Cuz when x is 9, you have to use this one. It includes 9. Now let's check the right side limit. As x approaches 9 from the right. So greater than nine we need to use this. So that's the square of x + 1. So the square root of 9 + 1. The square root of 9 is 3. 3 + 1 is 4. So the leftsided and the right sided limits they match. Therefore the limit as x approaches 9 of fx is equal to 4 and that also equals f of 9. So f of a is defined. We see that here. The limit exists and it equals f of a. Therefore f ofx is continuous at x= 9. And if it's continuous at x= 9, it does not have a discontinuity at x= 9. So the correct answer is answer choice D.