Checking for Continuous function

Question

I have the below question, i understand how to check for continuity for individual equations in a piecewise function but i dont understand how to find continuity for the whole function with only 1 inequalitygiven, could someone explain

Answer 1

The functions $f(x)=400x$ and $g(x)=600x-0.2x^2$ are continuous functions because they are polynomial functions (polynomial functions are continuous everywhere). So, the function $P(x)$ is continuous everywhere on its domain except that we should check the point where those two functions, $f(x)$ and $g(x)$, meet (if they don't meet, there will be some sort of gap or hole at the point where they're supposed to meet). If they meet at the point $x=1000$ (the only point in the domain of $P(x)$ that's most likely to make it discontinuous), then the function $P(x)$ is going to be continuous at that point and thus everywhere on its domain. So, if the following equality is true, $P(x)$ is continuous over $[500,2000]$ (otherwise it's going to be discontinuous over $[500,2000]$):

$$ f(1000)=g(1000)\implies\\ 400\cdot 1000 = 600 \cdot 1000 - 0.2 \cdot 1000^2\implies\\ 400000=400000\implies true $$

Therefore, $P(x)$ is continuous over $[500,2000]$.

Answer 2

How is this different from the piece-wise functions you know how to check?

$P(x)$ is continuous over the interval $(500,1000)$ because polynomials are continuous.

Same thing for the interval $(1000,2000)$

That leaves one point left to check.

$P(x)$ is continuous if

$\lim_\limits{x\to 1000^-} P(x) = \lim_\limits{x\to 1000^+} P(x) = P(1000)$

Answer 3

It is indeed, continuous. It is easy to check that the function is continous for each of the partitions [500,1000] and (1000, 2000). The "trouble" part is where x = 1000. You can see that 400(1000) = 400,000 = 600*(1000) - .2(1000)^2. The RHS of that expression is precisely the limit of the second part when iot approaches 1000 by the right. (sorry for not posting in LateX).