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

Question

Find all point of discontinuity of the function f(t) = \(\frac{1}{t^2+t-2}\), where t = \(\frac{1}{x-1}.\)

Answer 1

Clearly,

  t = \(\frac{1}{x-1}\) is discontinuous at x = 1 as t is not defined at this point.

∴ f(t) being a composition of function involving t, it is also discontinuous at x = 1

∵ f(t) = \(\frac{1}{t^2+t-2}\) 

= \(\frac{1}{(t+2)(t-1)}\)

by observing f(t) we can say that f(t) is not defined at 

t = – 2 and t = 1. 

∴ this will also contribute to discontinuity 

Hence,

t = – 2 = \(\frac{1}{x-1}\)

⇒ x – 1 = –1/2 

⇒ x = 1 – 1/2 

⇒ x = 1/2

Also other point of discontinuity is obtained by :

 t = 1 = \(\frac{1}{x-1}\)

⇒ x – 1 = 1 

⇒ x = 2 

Hence points at which f(t) is discontinuous are x = { 1/2,1,2} 

At all other point it is continuous.