I was wondering what the best way to solve questions like these are? $$\frac{x^3+3}{x^2+1}>\frac{x^3-3}{x^2-1}$$ I can get the answer, which is $(-\infty,-1)\cup(1,3)$. But I'm not sure if I have the most efficient method for the final steps. If this didn't involve fractions I'm quite happy sketching a cubic graph and choosing the correct sections. However with the fractional questions, I'm not sure how/if to relate the graph with answer - due to some x values being undefined and if you multiply by a negative then the inequality sign flips.
So far the way I do it is:
- rearrange until I can solve the cubic - this gives me x=0 as a repeated root and x=3
- look at the denominators to see where x cannot be defined - this tells me x$\neq$1 and x$\neq$-1
- list each seperate interval and test them all in the original inequality:
a) $(-\infty,-1)$ trying x = -2 in original inequality tells me this interval works
b) $(-1,0)$ try x=-0.5 this interval does not work
c) $(0,1)$ try x=0.5 this does not work
d) $(1,3)$ try x=2 this works
e) $(3,\infty)$ try x=4 this does not work
So the final answer is: $(-\infty,-1)\cup(1,3)$
Is this the only way to go? Or am I missing something? Any help is much appreciated
I suggest to proceed as follows
$$\frac{x^3+3}{x^2+1}>\frac{x^3-3}{x^2-1}$$
$$\frac{x^3+3}{x^2+1}-\frac{x^3-3}{x^2-1}>0$$
$$\frac{(x^3+3)(x^2-1)-(x^3-3)(x^2+1)}{(x^2+1)(x^2-1)}>0$$
$$\frac{2x^2(x-3)}{(x^2+1)(x^2-1)}<0$$
and from here study the sign of the terms $x-3$, and $x^2-1$ in the relavant intervals $(-\infty,-1)$, $(-1, 0)\cup(0, 1)$, $(1,3)$, $(3, \infty)$ which leads to the solution $x<-1$ or $1<x<3$.