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

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Answer 1 · 2016-12-13T06:34:05

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

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

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