I don't understand why this is not continuous.
The function is continuous everywhere except at x = 2. At x = 2, the limit from the left is 3 whereas the limit from the right is 2. So the function is not continuous at x = 2.
Please explain how to do this problem
Since the two parts of the function are defined with polynomials, the function is continuous for all x except possibly for x = 4. In order for the function to be continuous at x = 4, the limit from the left must be equal to the limit from the right. The limit from the left is the limit as x approaches 4 of the function 1 - 3x; so the limit from the left is 1 - 3(4) = -11. The limit from the right of f(x) at x = 4 is the limit as x approaches 4 of Ax^2 + 2x - 3; this limit is 16A + 8 - 3 = 16A + 5. The two limits are equal when -11 = 16A +5. Solving for A gives A = 2.
when I did this one, I got f(0)=-1 and f(1)=1. Is this right? The book had opposite answers. Also, how would you like the answer to a problem like this stated on the test? Do we just work the problem out, or do you also want it written out?
This is an application of the Intermediate Value Theorem. The function f(x) = cuberoot(x) - (x^2+2x-1) is continuous on the interval [0,1]. Also, f(0) = cuberoot(0) - (0^2+2(0)-1) = 1 and f(1) = cuberoot(1) - (1^2+2(1)-1) = -1. So, by the intermediate value theorem, there f(x) = d has a solution in [0, 1] for every d between -1 and 1. Letting d = 0, we see that there is a solution of f(x) = 0. This is the x we are looking for in the problem.
So, in answer to your question, the values of f at the end points agree with answer in the back of the book -- maybe your definition of f was different? The rest of the answer in the back of the book was a bit garbled.