Solving the inequality $\sqrt{1-|x|} \leq x-1$

Question

I am having trouble solving the following inequality:

$$\sqrt{1-|x|} \leq x-1$$

I need to find the smallest solution to this inequality.

Here's what I have tried so far:

• I first squared both sides of the inequality to get rid of the square root. This gives me:

$$1-|x| \leq x^2 - 2x + 1$$

• I then simplified this expression to get:

$$|x| - x^2 + 2x \leq 0$$

• Next, I tried to use the fact that $|x| = \pm x$ depending on the sign of $x$. This gives me two cases:

Case 1: $x \geq 0$

In this case, we have $|x| = x$. Substituting this into the inequality, we get:

$$x - x^2 + 2x \leq 0$$

Simplifying this expression, we get:

$$x(x-3) \leq 0$$

This means that $0 \leq x \leq 3$.

Case 2: $x < 0$

In this case, we have $|x| = -x$. Substituting this into the inequality, we get:

$$-x - x^2 + 2x \leq 0$$

Simplifying this expression, we get:

$$x(x+1) \leq 0$$

This means that $-1 \leq x \leq 0$.

• Combining the two cases, we get that the solution to the inequality is $-1 \leq x \leq 3$.

However, I am not sure if this is the smallest solution to the inequality. Can someone please check my work and let me know if I am on the right track?

Thank you!

Answer 1

Since $x-1<0$ is not possible then, there is only one case $x-1≥0$, and when squaring you need the restrictions $x-1≥0$ and $1-|x|≥0$ . But, we don't need the squaring operation. Because,

$$\begin{cases}x≥1\\ |x|≤1\end{cases}\implies x=1\thinspace .$$

This implies that, if there is a solution, then $x=1$ is an only possible root and we see that $x=1$ is a solution .

Answer 2

Maybe the easiest and simplest way is @lone_student's idea.

Your way will also work, but your attempt includes a mistake.

You deduce $|x|-x^2+2x\leqq 0$ from $1-|x|\leqq x^2-2x+1$, but this should be $-|x|-x^2+2x\leqq 0$.

Then separate the case into Case 1: $x\geqq 0$ and Case 2: $x<0$. You will get $x\geqq 1$ from Case 1 and $x<0$ from Case 2.

And, as @lost_student says, you should note $1-|x|\geqq 0$ and $x-1\geqq 0$, since the given problem is $\sqrt{1-|x|}\geqq x-1$ : If $1-|x|<0$, then $\sqrt{1-|x|}$ is not defined thus $1-|x|\geqq 0$ is necessary. Moreover, $\sqrt{\quad}$ is nonnegative thus it follows that $0\leqq\sqrt{1-|x|}\leqq x-1$ so $0\leqq x-1$ is necessary. From this , you get $-1\leqq x\leqq 1$ and $x\geqq 1.$

Finally, combining each case, you will get $$(x\geqq 1\quad\mathrm{or}\quad x<0)\quad\mathrm{and}\quad -1\leqq x\leqq 1\quad\mathrm{and}\quad x\geqq 1.$$ This is equivalent to $x=1.$

Answer 3

As noticed we need $x\ge 1$ and this leads to the solution.

Another way to proceed is by cases

• Case $1$: $x\ge 0$

$$\sqrt{1-|x|} \leq x-1 \iff \sqrt{1-x} \leq x-1$$

which requires

$$1-x\ge 0 \; \land \; x-1\ge 0 \iff x=1$$

• Case $2$: $x< 0$

$$\sqrt{1-|x|} \leq x-1 \iff \sqrt{1+x} \leq x-1$$

which requires

$$1+x\ge 0 \; \land \; x-1\ge 0 \iff x\ge 1$$

which is impossible.

Here a graphical sketch for the inequality

Answer 4

We can also apply properties of the functions involved in the inequality. The function $ \ y \ = \ \sqrt{1 - |x|} \ $ is the composition of $ \ y \ = \ \sqrt{1 - x} \ $ upon the even function $ \ y \ = \ |x| \ \ , \ $ which makes $ \ \sqrt{1 - |x|} \ $ even as well. So we can just look at how this function behaves for $ \ x \ \ge \ 0 \ \ . $

Since we are taking $ \ x \ $ to be a real variable, $ \ \sqrt{1 - x} \ $ is only defined for $ \ 1 - x \ \ge \ 0 \ \Rightarrow \ 1 \ \ge \ x \ \ . \ $ The even symmetry of the function $ \ \sqrt{1 - |x|} \ $ tells us that its domain is then $ \ -1 \ \le \ x \ \le \ 1 \ \ . \ $ As for the range, in the interval $ \ 0 \ \le \ x \ \le \ 1 \ \ , \ \ 1 \ \ge \ 1 - x \ \ge \ 0 \ \ , \ $ and so $ \ 1 \ \ge \ y \ = \ \sqrt{1 - x} \ \ge \ 0 \ \ ; \ $ applying even symmetry again, $ \ 1 \ \ge \ y \ = \ \sqrt{1 - |x|} \ \ge \ 0 \ \ . $

The second function is the linear $ \ y \ = \ x - 1 \ \ , \ $ which is always increasing (with slope 1) and has the $ \ y-$intercept $ \ (0 , -1) \ . \ $ In the domain interval $ \ -1 \ \le \ x \ \le \ 1 \ \ , \ \ -2 \ \le \ x - 1 \ \le \ 0 \ \ , \ $ and is only non-negative at $ \ (1 \ , \ 0) \ \ . \ $ Thus, $ \ \sqrt{1 - |x|} \ > \ x - 1 \ $ for $ \ -1 \ \le \ x \ < \ 1 \ \ $ and the two functions are only equal at $ \ \mathbf{ x \ = \ 1 }\ \ . \ $ This is the sole solution of the (improper) inequality $ \ \sqrt{1 - |x|} \ \le \ x - 1 \ $. (user presents a graph of the functions.)