# inequality

• Sep 6th 2008, 03:52 PM
Musab
inequality
Find the set of real values of x for which :

$\displaystyle \frac{2}{x} \geq 3-x$

I got x=-3 , x=1

what should I do next ?
• Sep 6th 2008, 04:05 PM
mr fantastic
Quote:

Originally Posted by Musab
Find the set of real values of x for which :

$\displaystyle \frac{2}{x} \geq 3-x$

I got x=-3 , x=1

what should I do next ?

Solve the following cases:

Case 1: $\displaystyle x > 0$: $\displaystyle 2 \geq 3x - x^2 \Rightarrow x^2 - 3x + 2 \geq 0$.

Case 2: $\displaystyle x < 0$: $\displaystyle 2 \leq 3x - x^2 \Rightarrow x^2 - 3x + 2 \leq 0$.

Edit: Note that $\displaystyle x \neq 0$ (why?)
• Sep 6th 2008, 04:41 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Solve the following cases:

Case 1: $\displaystyle x > 0$: $\displaystyle 2 \geq 3x - x^2 \Rightarrow x^2 - 3x + 2 \geq 0$.

Case 2: $\displaystyle x < 0$: $\displaystyle 2 \leq 3x - x^2 \Rightarrow x^2 - 3x + 2 \leq 0$.

Case 1: $\displaystyle x^2 - 3x + 2 \geq 0 \Rightarrow$ $\displaystyle x \leq 1$ or $\displaystyle x \geq 2$ subject to the restriction $\displaystyle x > 0$. Therefore $\displaystyle x \geq 2$ or $\displaystyle 0 < x \leq 1$.

Case 2: Left for you.