Find the set of values of x for which (x+4) > 2/(x+3)

This is an example of an inequalities question from FP2. For this, we will need to use the tools learned in this chapter. To start with, it may be tempting to multiply both sides of the inequality by (x+3) to get rid of the fraction, but doing this is wrong since in the case that (x+3) is negative (when x < -3), the direction of the inequality will not be preserved. Hence, we proceed by multiplying both sides by (x+3)² (which is always non-negative). We then arrive at (x+3)²(x+4) > 2(x+3). Using algebraic rearrangement and factorisation we can then get to (x+3)[(x+3)(x+4)-2] > 0. This is a good place to get to, since we can see that there is a quadratic (which we can factorise) in the second term. Expanding this out we reach (x+3)(x²+7x+10) > 0. Now we can factorise the quadratic (we find 2 numbers 5 and 2 that add to 7 and multiply to 10) to get (x+3)(x+5)(x+2) > 0. We can clearly see this is a cubic expression on the left hand side. Now we can draw the graph y = (x+3)(x+5)(x+2), which must intersect the x axis at x = -5, -3 and -2 (since these value of x give a y value of 0). Now, looking at the annotated graph, we can see that the desired region (where y < 0) must be where x > -2 or -5 < x < -3. Note that we use strict inequality here and not equality aswell since if x were eqeal to these values, y would be equal to 0, which is outside of the constraint.