okay, so i need to find:
A) for which values of x is x^2 + x + 1 >= (x-1)/(2x-1)
and another similar one
for which values of x is -3x^2 +4x > 1?
Mmmk.
A) $\displaystyle x^2+x+1 \ge \frac{x-1}{2x-1}$
Get rid of the fraction on the RHS. Multiply through by 2x-1.
$\displaystyle (2x-1)[x^2+x+1] \ge x-1$
$\displaystyle 2x^3-x^2+2x^2-x+2x-1 \ge x-1$
$\displaystyle 2x^3+x^2 \ge 0$
$\displaystyle 2x+1 \ge 0 $
Can you take it from here?
(Sorry if I messed up the algebra. It's late and I did it in my head, but you get the idea I hope)
Hello,
for confirmation only I give you the solution of the first one:
$\displaystyle x=0~\vee~x \leq-\frac{1}{2}~\vee~x>\frac{1}{2}$
#2.
$\displaystyle -3x^2 +4x > 1~\Longrightarrow~-3x^2 +4x -1> 0$. Now factor the LHS of the inequaltiy:
$\displaystyle (-3x+1)(x-1)>0$. You have a product of 2 factors which is positiv. That means both factors must have the same sign. The sign + means that the value is greater than zero, the sign - ...:
$\displaystyle -3x+1>0\ \wedge \ x-1>0~\vee~-3x+1<0\ \wedge \ x-1<0$. Solve each inequality:
$\displaystyle x<\frac{1}{3}\ \wedge \ x>1~\vee~x>\frac{1}{3}\ \wedge \ x<1$. Collect the results of inequalities which are connected by $\displaystyle \wedge$:
$\displaystyle \emptyset\ \vee\ \frac{1}{3} <x< 1$
$\displaystyle \displaystyle x^2+x+1\geq\frac{x-1}{2x-1}\Leftrightarrow\frac{(x^2+x+1)(2x-1)-x+1}{2x-1}\geq 0\Leftrightarrow\frac{x^2(2x+1)}{2x-1}\geq 0$
Now we make the sign of each factor in a table and we use the rule of signs.
$\displaystyle \begin{tabular*}{0.75\textwidth}{ | c | c c c c c |}
\hline
$x$ & $-\infty$ & $-\frac{1}{2}$ & $0$ & $\frac{1}{2}$ & $\infty$\\
\hline % put a line under headers
$x^2$ & + & + & 0 & + & +\\
\hline
$2x+1$ & $-$ & $0$ & $+$ & $+$ & $+$ \\
\hline
$2x-1$ & $-$ & $-$ & $-$ & 0 & $+$\\
\hline
$E(x)$ & + & 0 & $-0-$ & $|$ & $+$\\
\hline
\end{tabular*}$
From the last row in the table we have
$\displaystyle x\in\left(\left.-\infty,-\frac{1}{2}\right]\right.\cup\{0\}\cup\left[\left.\frac{1}{2},\infty\right)\right.$