Should $x=2$ be included in the solution?

Question

Consider an inequation $\frac{(x-2)^6 (x-3)^3 (x-1)}{(x-2)^5 (x-4)^3}>0$. $(x-2)^5$ can be cancelled from both the numerator and denominator with the condition that $x\neq2$ leaving $(x-2)$ in the numerator. Since $(x-2)$ is still in the numerator, , it is a critical point and has to be marked on the number line while solving the inequation using wavy curve method. Should $x=2$ be included in the solution or not due to condition of cancellation?

Answer 1

Your function has a removable discontinuity at $x=2$ - the function is not defined there.

If you remove the discontinuity, the value is $0$ and thus the point would not be part of the solution set anyway.

Removing the discontinuity could help in analysing the problem because the function changes sign from $x\lt 2$ to $x\gt 2$. The change of sign exists whether you remove the discontinuity or not, and is relevant to analysis of the solution, but the theorems which go with continuous functions are available to you if you remove the discontinuity. (Effectively one point is removed from the graph.)

Note that if you had the same exponent in numerator and denominator, removing the discontinuity would give a non-zero value, and you would have to exclude the point from the solution set.