Question

How do solve `(x+1)^2/(x^2+2x-3)<=0` and write the answer as a inequality and interval notation?

Answer 1

The answer is `-3< x <1`
or `x in ]-3,1[`

Explanation:

Let's factorise the denominator

`x^2+2x-3=(x-1)(x+3)`

Let `f(x)=(x+1)^2/((x-1)(x+3))`

The domain of `f(x)` is `D_f(x)=RR-{-3,1}`

The numerator is `>0, AA x in D_f(x)`

Now, we can do the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+`

So,

`f(x)>=0` when `x in ] -3,1 [` or `-3< x < 1`

Let `graph{y=(x+1)^2/((x-1)(x+3)) [-12.66, 12.65, -6.33, 6.33]}`