How do solve $\frac{x^{2} - x - 12}{x^{2} + 4} \leq 0$ $(x^2-x-12)/(x^2+4)<=0$ and write the answer as a inequality and interval notation?

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Answer 1 · 2017-01-21T17:56:08

The answer is $x \in [- 3, 4]$ $x in [-3, 4 ]$ or $- 3 \leq x \leq 4$ $-3<=x<=4$

Let's factorise the numerator

$x^{2} - x - 12 = (x + 3) (x - 4)$ $x^2-x-12=(x+3)(x-4)$

Let $f (x) = \frac{x^{2} - x - 12}{x^{2} + 4}$ $f(x)=(x^2-x-12)/(x^2+4)$

The domain of $f (x)$ $f(x)$ is $RR$

The denominator is $>0$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaa)$ $-oo$ $color(white)(aaaa)$ $-3$ $color(white)(aaaa)$ $4$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $x+3$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-4$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<=0$ when $x in [-3, 4 ]$

$-3<=x<=4$

How do solve (x^2-x-12)/(x^2+4)<=0x2−x−12x2+4≤0(x^2-x-12)/(x^2+4)<=0 and write the answer as a inequality and interval notation?