Define a piece-wise function f(x) that has a removable discontinuity at x=6, is only left continuous at x=2, and has no continuity at x=-3

Question

Define a piece-wise function f(x) that has a removable discontinuity at x=6, is only left continuous at x=2, and has no continuity at x=-3

Answer 1

f(x) = x / (x²-x-12), if x ≤ 2 [vertical asymptote when x = -3 and f(2) = -1/5]

= (x²-7x+6) / (x²-8x+12), if x > 2 [removable discontinuity at x = 6(hole in graph) and as x approaches 2 from the right, f(x) approaches ∞]

Answer 2

We can define a piecewise function that satisfies the conditions as follows:

1. Removable discontinuity at : This means the limit exists at , but is not equal to the limit. We can create this by defining a function that is continuous everywhere except at , and then redefine to create the discontinuity.

2. Left continuous at : The function is continuous from the left at (i.e., ), but it is not continuous from the right (i.e., ).

3. No continuity at : This can be achieved by defining a jump discontinuity or any other type of discontinuity at .

Here's a piecewise definition for that meets thes

e criteria: