1. ## function problem

Hello! I have this problem.

$f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}$

2. Originally Posted by Anemori
Hello! I have this problem.

$f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}$
Solve $\frac {x^2}{x^2+1} = \frac {1}{2}$ .... (1)

Then draw a graph of $y = \frac {x^2}{x^2+1} = - \frac{1}{x^2 + 1} + 1$ and use it and the solutions to equation (1) above to solve the given inequation.

3. Originally Posted by Anemori
Hello! I have this problem.

$f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}$
Here is a slightly different approach:

Since $x^2 + 1 \geq 1$ the inequality becomes:

$x^2\geq\frac12 x^2 + \frac12$

$\frac12 x^2- \frac12\geq 0~\implies~x^2-1\geq 0~\implies~(x+1)(x-1)\geq 0$

A product of 2 factors is positive if both factors have the same sign:

$x+1\geq 0\wedge x-1\geq 0~\vee ~x+1\leq 0\wedge x-1\leq 0$

$x\geq -1\wedge x\geq 1~\vee ~x\leq -1\wedge x\leq 1$

$x\geq 1~\vee ~x\leq -1$

4. Originally Posted by Anemori
Hello! I have this problem.

$f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}$
Rewrite it as $\frac{x^2}{x^2+1}- \frac{1}{2}= \frac{2x^2}{2(x^2+1}- \frac{x^2+ 1}{2(x^2+ 1}= \frac{x^2- 1}{x^2+ 1}\ge 0$

A fraction is positive if and only if the numerator and denominator have the same sign. This, denominator, $x^2+ 1$, is always positive.