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Math Help - inequality/proof type thing

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by mistykz View Post
    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)
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  3. #3
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    Quote Originally Posted by mistykz View Post
    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} <x< 1
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  4. #4
    MHF Contributor red_dog's Avatar
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    \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 |}<br />
\hline<br />
$x$ & $-\infty$ & $-\frac{1}{2}$ & $0$ & $\frac{1}{2}$ & $\infty$\\<br />
\hline % put a line under headers<br />
$x^2$ & + & + & 0 & + & +\\<br />
\hline<br />
$2x+1$ & $-$ & $0$ & $+$ & $+$ & $+$ \\<br />
\hline<br />
$2x-1$ & $-$ & $-$ & $-$ & 0 & $+$\\<br />
\hline<br />
$E(x)$ & + & 0 & $-0-$ & $|$ & $+$\\<br />
\hline<br />
\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.
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  5. #5
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    thanks

    awesome, thanks everyone!
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