I'm trying to solve an exercise and I'm having troubles to get it right.
We have two functions $f(x)$ and $g(x)$ defined as follow :
- $f(x) = 2x-3$
- $g(x) = -x²+x-3$
Question : We have to demonstrate that solving $f(x) > g(x)$ is equivalent to solve $x(-x-1)>0$. It is not allowed here to use the polynomial form to solve the inequalities.
My researches : I first draw the representative curves of the functions and observed that :
On $]-\infty;-1]$ we have $f(x) > g(x)$
On $]-1 ; 0]$ we have $f(x) < g(x)$
On $]0 ; +\infty[$ we have $f(x) > g(x)$
Then, I tried to develop the inequality $f(x) > g(x)$ as follow:
$f(x) > g(x)$
$2x-3 > -x²+x-3$
$2x-3+x²-x+3 > 0$
$x² + x >0$
$x(x+1) > 0$
The two roots here are obviously $x=0$ and $x=-1$ and for the inequality to be true we must have either $x>0$ or $x<-1$.
However for $x(-x-1) > 0$ to be true we must have $-1<x<0$.
Therefore solving $f(x) > g(x)$ is NOT equivalent to solve $x(-x-1)>0$.
Am I getting something wrong here ? Or is there a mistake in the exercise ?
Thank you very much for reading this post (that may sound quite silly...).
I think it is $$x<-1$$ or $$x>0$$