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I tried to figure out this question. Can any one help? I know the answer but I keep getting the wrong one. Thanks in advance.

Find all values of x that satisfy $|-(x + 1)^2+1|\geq 1$

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I have retagged your question. The tag (algebraic-geometry) is intended for questions in the field of Algebraic Geometry, a relatively advanced topic usually reserved for graduate courses. –  Alex Becker Jul 11 '12 at 3:12
    
For $M > 0, |x| \ge M \iff x \le -M$ or $x \ge M$ –  Vectk Jul 11 '12 at 3:19
    
First, you can remove the minus sign, as $|x|=|-x|$. –  Ross Millikan Jul 11 '12 at 3:20

6 Answers 6

$|x|\geq a\implies x\geq a$ or $x\leq -a$.Here, in your problem it results to, $1-(x+1)^2\geq 1\implies (x+1)^2\leq 0$, only solution for which is $x=-1.$ Also, there is a second case, $1-(x+1)^2\leq -1\implies (x+1)^2\geq 2\implies |x+1|\geq\sqrt 2\implies x\geq\sqrt 2-1$ or $x\leq -\sqrt2-1$.Therefore, the possible solutions of $x$ are $\{-1\}\cup(-\infty,-\sqrt2-1]\cup[\sqrt2-1,\infty)$ .

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How did you get from (x+1)^2>=0 to x = -1. I thought it would be x >= -1 –  Mark Jul 11 '12 at 3:27
    
That is not $(x+1)^2\geq 0$ but $(x+1)^2\leq 0$ and since $(x+1)^2$ is a square, it can't be negative, hence $(x+1)^2=0\implies x=-1$. –  Aang Jul 11 '12 at 5:02

You can observe that $y=-(x+1)^2+1$ is a parabola with vertex in $(-1,1)$ and $a=-1$. Then you can draw its absolute value, and the line $y=1$.

If you solve the equation $(x+1)^2-1=1$ you will find the two intersection wich are, respectively, less than $-2$ and bigger than $0$ (i.e., $-1\pm\sqrt2$).

Finally you can write the solution: $x\le -1-\sqrt2 \vee x=-1 \vee x\ge-1+\sqrt2$.

enter image description here

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1  
$x=1$ is not the solution, $x=-1$ is. –  Aang Jul 11 '12 at 9:16
    
uh, right, I have fixed it, thank you –  zar Jul 11 '12 at 9:24

$|b-a|$ is the distance between $b$ and $a$ on the real line. So you look for $(x+1)^2 \leq 0 $ or $(x+1)^2\geq 2$.

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$$|-(x + 1)^2+1|\geq 1 >0$$

Since both sides of the inequlity are positive squaring the inequality we get,

$$((x + 1)^2-1)^2 \geq 1 $$

$$\Leftrightarrow ((x + 1)^2-1)^2-1^2\geq 0 \tag {difference of squares} $$

$$\Leftrightarrow ((x + 1)^2-1-1)((x + 1)^2-1+1)\geq 0 $$

$$\Leftrightarrow ((x + 1)^2-2)(x+1)^2 \geq 0 \tag {with $(x+1)^2>0$ iff $x \neq -1$} $$

Hence,

Case 1 $x = -1$

Then, $((x + 1)^2-2)(x+1)^2 =0 \geq 0$ as required

Case 2 $x \neq -1$

Then $(x + 1)^2-2\geq 0$ (dividing the inequality by $(x + 1)^2>0$)

$$\Leftrightarrow(x + 1- \sqrt{2})(x + 1+ \sqrt{2})\geq 0$$

$$ \Leftrightarrow x \leq -1-\sqrt{2} \ or \ x \geq -1+\sqrt{2}$$

Therefore the final solution is $$x \in (- \infty,-1-\sqrt{2} ] \cup \{-1 \} \cup [-1+\sqrt{2}, +\infty)$$

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A very slightly different approach to the ones already given is to recall that $|y| = \sqrt{y^2}$. So we need to find $x \in \mathbb{R}$ such that

$$ \sqrt{\left(1-(x+1)^2\right)^2} \geq 1 $$

Squaring both sides you get

$$ \left(1-(x+1)^2\right)^2 \geq 1 \quad \Longrightarrow \quad -2(x+1)^2 + (x+1)^4 \geq 0 $$

If $x \neq -1$ (check that it is also a solution), we can divide both sides by $(x+1)^2$ and fall onto a quadratic equation that is easy to solve. Finally, the solution set is $$ \left\lbrace x \in \mathbb{R} \;|\; x \leq -1-\sqrt{2} \;\;\text{or}\;\; x=-1 \;\;\text{or}\;\; x \geq -1+\sqrt{2} \right\rbrace $$

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Observe that $(x+1)^2\geq 0$, so $1-(x+1)^2\leq 1$, with equality if and only if $(x+1)^2=0$ if and only if $x+1=0$ if and only if $x=-1$. Thus, $x=-1$ is a solution to the inequality.

The only other option to satisfy it is when $1-(x+1)^2\leq -1$ if and only if $-x^2-2x\leq -1$ if and only if $0\leq x^2+2x-1$. Starting with equality, we find from the quadratic equation that $0=x^2+2x-1$ if and only if $x=-1\pm\sqrt{2}$. Using test points in the intervals $(-\infty,-1-\sqrt{2})$, $(-1-\sqrt{2},-1+\sqrt{2})$, $(-1+\sqrt{2},\infty)$, it is readily seen that $x^2+2x-1$ is positive on the first and last interval and negative on the middle. Thus, $x^2+2x-1\geq 0$ if and only if $x\leq -1-\sqrt{2}$ or $x\geq -1+\sqrt{2}$.

Consequently, the solution set is $\left(-\infty,-1-\sqrt{2}\right]\cup\{-1\}\cup\left[-1+\sqrt{2},\infty\right)$.

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the points $-1-\sqrt 2$ and $-1+\sqrt 2$ should be included in solution set as they satisfy the given inequality. –  Aang Jul 11 '12 at 5:01
    
D'oh! Closed rays. Thanks. –  Cameron Buie Jul 11 '12 at 13:51

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