Precalculus continuity problem, use the 3 step definition of continuity to discuss the continuity (question is for jimthompson5910)

As the name implies, there are 3 parts or conditions that must be held true in order for the function to be continuous at x = a.

Those three conditions are:

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In this case, a = -1 as this is where the junction happens. It's where f(x) swaps identity from the first piece x^2-3 to the second piece 3x+1

To compute f(a), aka f(-1), we'll use the second piece

f(x) = 3x+1

f(-1) = 3(-1)+1

f(-1) = -2

We see that f(a) does indeed exist, so condition (1) is held true.

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To compute the limit, we'll need the left hand limit (LHL) and right hand limit (RHL)

The LHL is found by having x approach -1 from the left. So you'll start with something like x = -2, then move to x = -1.5 then to x = -1.1 then to x = -1.01, and so on, getting steadily closer to x = -1. We won't actually arrive at x = -1 itself.

But because x^2-3 is a polynomial, and all polynomials are continuous, we can simply plug x = -1 into this to find that...

y = x^2-3

y = (-1)^2 - 3

y = -2

The input x = -1 leads to the output y = -2 for the first piece. This is the LHL.

We found earlier that x = -1 lead to y = -2 for the second piece. This is the RHL.

Because LHL = RHL, we have proven condition (2) is true. It also means condition (3) is true because f(a) is part of the RHL, more or less.

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In layman's terms, we can think of the two curves as roads. For the function to be continuous, the road cannot have any jumps, gaps, or potholes. Condition (1) says that the point must exist, aka there isn't a pothole there. Condition (2) says that the two pieces of the road, on either side of the point in question, must connect together. Hence there are no gaps or jumps. Condition (3) effectively ties everything together.

You might be asking if condition (3) is a bit redundant. Surely if f(a) exists and the limit exists, then that's enough to prove continuity, right? Unfortunately no that's not the case. Consider the situation where the limit exists, but f(a) was some other value. That means we have a removable discontinuity. That point in the road is a pothole but the two roads do connect.

One could argue that conditions (2) and (3) are sufficient, and condition (1) isn't really needed. This is because if condition (3) was the case, then we automatically have shown that f(a) must exist. This is of course assuming we found that the limit exists as well, and it's not plus/minus infinity.

congruent segments

step-by-step explanation: