Solve: $|-(x + 1)^2+1|\geq 1$

Question

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

Answer 1

$|x|\geq a\implies x\geq a$ 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 $\{-1\}\cup(-\infty,-\sqrt2-1]\cup[\sqrt2-1,\infty)$ .

Answer 2

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: $x\le -1-\sqrt2 \vee x=-1 \vee x\ge-1+\sqrt2$.

Answer 3

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

Answer 4

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

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 $$x \in (- \infty,-1-\sqrt{2} ] \cup \{-1 \} \cup [-1+\sqrt{2}, +\infty)$$

Answer 5

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 $$ \left\lbrace x \in \mathbb{R} \;|\; x \leq -1-\sqrt{2} \;\;\text{or}\;\; x=-1 \;\;\text{or}\;\; x \geq -1+\sqrt{2} \right\rbrace $$

Answer 6

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 $\left(-\infty,-1-\sqrt{2}\right]\cup\{-1\}\cup\left[-1+\sqrt{2},\infty\right)$.