Question

Find all points of discontinuity of the function f(t) = `1/("t"^2 + "t" - 2)`, where t = `1/(x - 1)`

Answer 1

We have, f(t) = `1/("t"^2 + "t" - 2)`

Where t = `1/(x - 1)`

∴ f(t) = `1/((1/(x - 1))^2 + 1/(x - 1) - 2)`

= `(x - 1)^2/(1 + (x - 1) - 2(x - 1)^2)`

= `(x - 1)^2/(-(2x^2 - 5x + 2))`

= `(x - 1)^2/((2x - 1)(2 - x))`

So, f(t) is discontinuous at 2x – 1 = 0

⇒ x = `1/2` and 2 – x = 0

⇒ x = 2

Also f(t) is discontinuous at x = 1, where t = `1/(x - 1)` is discontinuous.