Please solve
(x^2)-4/x-4 >0
in order for the fraction to be > 0, the sign of the numerator must agree with the denominator, that is, if the numerator is positive, the denominator is positive as well, and if the numerator is negative, so is the denominator.
assume without loss of generality that both are positive (that means a similar thing happens if both are negative, so we don't have to consider that). so we have:
$\displaystyle x^2 - 4 > 0$ and $\displaystyle x - 4 > 0$
$\displaystyle \Rightarrow (x + 2)(x - 2)> 0$ and $\displaystyle x > 4$
$\displaystyle \Rightarrow x > -2$ or $\displaystyle x > 2$ and $\displaystyle x > 4$
so now, draw a number line and mark the values -2, 2 and 4. test all the regions between them and at the ends to see which region satisfies the original inequality
really? what about x = -1?
just test the regions as i said. plug in any value between the numbers that you marked on the number line into the original equation to see if they work, then from that, write your answer. the asnwer should be the union of two intervals, i believenow i graphed it its all simultaneous. but the x>4 one seems to be the most correct one? im dumb with this stuff lol.