# Thread: inequality/proof type thing

1. ## inequality/proof type thing

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?

2. Originally Posted by mistykz
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) $x^2+x+1 \ge \frac{x-1}{2x-1}$

Get rid of the fraction on the RHS. Multiply through by 2x-1.

$(2x-1)[x^2+x+1] \ge x-1$

$2x^3-x^2+2x^2-x+2x-1 \ge x-1$

$2x^3+x^2 \ge 0$

$2x+1 \ge 0$

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)

3. Originally Posted by mistykz
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?
Hello,

for confirmation only I give you the solution of the first one:

$x=0~\vee~x \leq-\frac{1}{2}~\vee~x>\frac{1}{2}$

#2.

$-3x^2 +4x > 1~\Longrightarrow~-3x^2 +4x -1> 0$. Now factor the LHS of the inequaltiy:

$(-3x+1)(x-1)>0$. 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 - ...:

$-3x+1>0\ \wedge \ x-1>0~\vee~-3x+1<0\ \wedge \ x-1<0$. Solve each inequality:

$x<\frac{1}{3}\ \wedge \ x>1~\vee~x>\frac{1}{3}\ \wedge \ x<1$. Collect the results of inequalities which are connected by $\wedge$:

$\emptyset\ \vee\ \frac{1}{3}

4. $\displaystyle x^2+x+1\geq\frac{x-1}{2x-1}\Leftrightarrow\frac{(x^2+x+1)(2x-1)-x+1}{2x-1}\geq 0\Leftrightarrow\frac{x^2(2x+1)}{2x-1}\geq 0$
Now we make the sign of each factor in a table and we use the rule of signs.

$\begin{tabular*}{0.75\textwidth}{ | c | c c c c c |}
\hline
x & -\infty & -\frac{1}{2} & 0 & \frac{1}{2} & \infty\\
\hline % put a line under headers
x^2 & + & + & 0 & + & +\\
\hline
2x+1 & - & 0 & + & + & + \\
\hline
2x-1 & - & - & - & 0 & +\\
\hline
E(x) & + & 0 & -0- & | & +\\
\hline
\end{tabular*}$

From the last row in the table we have
$x\in\left(\left.-\infty,-\frac{1}{2}\right]\right.\cup\{0\}\cup\left[\left.\frac{1}{2},\infty\right)\right.$

5. ## thanks

awesome, thanks everyone!