# Checking for Continuous function

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

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 $$[500,2000]$$.

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

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

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).