How to check if a wavefunction is continous or not? and how to find probability? How to find if the following function is continuous? how do I find its derivative? Also, how would I find the probability of finding the function at $x = a-2$ and $x = b+4$?
$$ \psi(x,0) =\frac{Ax}{a}, \ \ if \ 0 \leq x \leq a$$
$$ \psi(x,0) =\frac{A(b − x)}{b − a}, \ \ if \ a \leq x \leq b$$
$$ \psi(x,0) = 0, \ \ otherwise$$
 A: Both $\frac{Ax}{a}$ and $\frac{A(b-x)}{(b-a)}$ are polynomial functions which when graphed are functions that have no holes, no jumps or asymptotes and behave well with limits, etc. They are continuous functions on the real line. And for $0$ the function obviously only has one value over it’s domain so it is continuous as well. 
To find the derivative of this wavefunction, just take the derivatives of each of the seperate rules.
As for finding the probability of $x = a-2$ and $x = b+4$, you must remember that $ΨΨ^{*}$ is not a function that gives you the probability for a particle being at any one individual point in space, it is a probability density, which means you can find the probability that a particle exists within a certain region of space, but not any one individual point. Technically speaking, there are an infinite amount of points in space so the probability of a particle being at any one point is actually zero. Note that contrary to popular belief a probability of zero does not necessarily equal impossible. 
So if you want to find the probability for a particle being in one of those regions, simple integrate the probability density over that region.
A: $\Psi^*\Psi=|\Psi(x)|^2$ according to the Born Rule is the probability density function $P(x)$, not the 'point probability' in $x$.
The probability of finding the particle in a part $\delta x$ of the particle's domain is  given by (for a normalised wave function):
$$P(x, \delta x)=\int_x^{x+\delta x}\text{d}x|\Psi(x)|^2$$
It is obvious that for $\delta x=0$ $\Rightarrow P(x, \delta x)=0$.
So  the point probabilities in the two points you're looking at is effectively $0$.
It's important to note this as many textbooks lazily abbreviate probability density function to probability. They are NOT the same.
