Question

Let f be given by the formula?

Let f be given by the formula:

Find the value(s) of x at which f is discontinuous?

Answer 1

At `x=1`

Explanation:

Consider the denominator.
`x^2 + 2x -3`
Can be written as:
`x^2 + 2x +1 -4`
`(x + 1)^2 -4`
`(x + 1)^2 -2^2`

Now from relation `a^2-b^2` = `(a+b)(a-b)` we have
`(x+1 +2)(x+1 -2))`

`(x+3)(x-1))`

If `x=1`, the denominator in above function is zero and the function tends to `oo` and not differentiable. Is discontinous.

Answer 2

`f(x)=(x+2)/(x^2+2x-3)` is discontinuous when `x=-3` and `x=1`

Explanation:

`f(x)=(x+2)/(x^2+2x-3)` is discontinuous when denominator is zero i.e.

`x^2+2x-3=0`

or `x^2+3x-x-3=0`

or `x(x+3)-1(x+3)=0`

or `(x-1)(x+3)=0`

i.e. `x=-3` and `x=1`

graph{(x+2)/(x^2+2x-3) [-10, 10, -5, 5]}