I finished the first question with no problem. It was something like:
Find the points at which f(x) [insert three equations for left right and in between here] is discontinuous. At each of these points, is f cont. from the right or the left?
As I said, I solved that one. I am not quite sure how to start the next one.
I think I would use the Intermediate Value Theorem, but I am not sure. Plus, if that's the case, do I do as above with the first equation?
Find the value k so that f(x) is continuous from the left at x=2, for
f(x)= | x+1, if x<=1;
| k, if x=2;
| x^2-5, if x>2
One, I am unsure if this is an IVT usage, or not (It looks like one to me), and two, I am not sure how to start it.
To find the value of k so that f(x) is continuous from the left at x=2, we need to use the definition of continuity. The Intermediate Value Theorem may not be directly applicable in this case.
To determine if f(x) is continuous from the left at x=2, we need to check if the left-hand limit, f(2-), is equal to the value of f(2).
First, let's determine the left-hand limit of f(x) as x approaches 2 (or f(2-)).
Since this function is defined piecewise, we need to consider each part individually.
For x < 2:
f(x) = x+1
As x approaches 2 from the left, f(x) approaches 2+1 = 3.
Now, let's consider the value of f(2):
f(2) = k
For f to be continuous from the left at x=2, the left-hand limit (f(2-)) should be equal to f(2).
So, equating the two values we found:
3 = k
Therefore, the value of k that makes f(x) continuous from the left at x=2 is k = 3.
To summarize:
- We determined the left-hand limit, f(2-), by evaluating the function for x < 2.
- We compared f(2-) with f(2) to check for continuity from the left.
- We found that k = 3 is the value that satisfies this condition.
Please let me know if you have any further questions or if there is anything else I can help you with!