Question

How do you prove that the function: `T(x) = 1 / (abs(x-2)-x^2)` is continuous between [1.5,8]?

Answer 1

Only ( infinite ) discontinuities at `x = -2 and x = 1`. The vertical asymptotes are `x = 1 and x =-2`. The horizontal asymptote is x = 0. y-intercept is `1/2`.

Explanation:

Besides the graph, I have provided some data for clarity. Further, the

given function is representing the couple

`T=1/(x-2-x^2), x>=2` and

`T=1/(2-x-x^2), x<=2.`

graph{y(x^2-|x-2|)+1=0 [-10, 10, -5, 5]}