Andy C. answered • 11/14/17

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-x / (x+3) < x+4

-x/(x+3) - (x+4) < 0 <--- subtracts (x+4) from both sides

[-x - (x+4)(x+3)]/(x+3) < 0 <---- common denominator of (x+3)

[ -x - (x^2 + 7x + 12]/(x+3) < 0 <--- FOIL

[ -x -x^2 - 7x - 12 ] / (x+3) < 0 <--- distributive of -1 over the quadratic

[-x^2 - 8x - 12]/(x+3) < 0 <--- combines like terms

(x^2 + 8x + 12)/(x+3) > 0 <--- multiplies both sides by -1 switch the inequality

(x + 6)(x + 2 )/(x+3) > 0 <--- factors

Not one single step involves multiplying by

a potential negative number. It was avoided

like the plague!

Let f(x) = (x+5)(x+2)/(x+3)

We are interested in intervals of x, where

f(x) >0.

There is a vertical asymptope at x=-3.

The zeros are x=-2, and x=-5

The intervals are x<-5, -5 < x < -3, -3 < x < -2, x>-2

Now it's time for the sign table. Just to make double sure,

The last row is the ORIGINAL problem -x/(x+3) < x+4

Interval (x < -5) (-5 < x < -3) (-3 < x < -2) (x>-2)

------------------------------------------------------------------------

Test X -10 -4 -2.5 0

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sign of f(x) - - - + - - + - + + + +

---------------------------------------------------------------------------

result - + - +

------------------------------------------------------------------------

Original problem -10/7<-6 -4 < 0 5 <1.5 0 < 4

FALSE TRUE FALSE

Notice that whenever f(x) is positive, the

original problem results in a true statement.

So (-5 < x < -3) or (x>-2)