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Just wanted to check my maths:

a) $\displaystyle \frac{(x^3-4)}{x-2} \leq x+2$ I got that $x\leq 1$.

b) $\displaystyle \frac{(2x-3)(x+1)}{x^2} \leq -2$ I got that $\displaystyle -\frac{3}{4} \leq x \leq 1$

Right/wrong? Thanks :). I just worked on it as if it were an equals sign... so I'm not sure I'm correct.

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Dangerous to work here just as if it were an equal sign... –  1015 Feb 11 '13 at 20:19
    
Please explain how you did it, so we can tell you if you are on the right track or just got the right answer by chance. –  vonbrand Feb 11 '13 at 20:43
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2 Answers

Let us do it very carefully.

You want to solve the inequality $$\frac{x^3-4}{x-2}\le x+2.$$

$1.$ It would be nice to get rid of the denominator. If $x\gt 2$, then the denominator is positive. Multiply both sides by $x-2$. We get the equivalent inequality $x^3-4\le x^2-4$, which is equivalent to $x^3\le x^2$. This is never true for $x\gt 2$.

$2.$ If $x\lt 2$, the denominator is negative. So multiplying both sides by $x-2$ reverses the inequality. Thus we are solving $x^3-4\ge x^2-4$, or equivalently $x^3\ge x^2$. This can be rewritten as $x^2(x-1)\ge 0$. That gives the possibility $x=0$ or $x\ge 1$.

But recall that the calculation for Case $2$ was done under the assumption $x\lt 2$. So the solutions of our inequality are $x=0$ and $1\le x\lt 2$.

The second inequality was handled almost correctly: there is no reversal issue, since $x^2$ is never negative. However, the left-hand side is not defined at $x=0$, and therefore $0$ should not be part of your solution set.

Another way: The only places that the direction of the inequality could change are (1) the places where we have equality and (2) the places where there is a singularity. For your first problem, there is a singularity at $x=2$. Now find all the places where the inequality direction could change, and use suitable "test points." I imagine you know how to handle this method: if detail is needed it can be supplied.

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Thanks I completely understand you! Thanks for your effort and time :). –  user61871 Feb 11 '13 at 21:32
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In case of the 2nd one,

$\displaystyle \frac{(2x-3)(x+1)}{x^2} \leq -2$

Multiplying both sides by $x^2$ (Considering $x\ne0)$as it is +ve so no problem of sign changing,

$\Rightarrow(2x-3)(x+1)\leq -2x^2$

$\Rightarrow 2x^2-x-3\leq -2x^2$

$\Rightarrow 4x^2-x-3\leq 0$

$\Rightarrow(2x)^2-2.2x.\frac{1}{4}+1/16-1/16-3\leq0$

$\Rightarrow(2x-1/4)^2-49/16\le0$

$\Rightarrow-\frac{7}{4}\le2x-1/4\le7/4$

$\Rightarrow-3/4\le x\le 1$

This solution must be without 0.

$x\in(-3/4,1)-{0}$

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