Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$

Question

Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$

My working out so far:

$|x-1|$ =

$x-1$ when $x≥1$

$1-x$ when $x<1$

Then we compute $x^2 + |x-1|$ as follows:

$x^2 + |x-1|$ =

$x^2 +x -1$ when $x≥1$

$x^2 - x + 1$ when $x<1$

So we have the following 2 cases:

i) When $x≥1$, we get $x^2 + |x-1| = x^2 +x -1$ and substituting this into the inequality $x^2 + |x-1| ≤ 1$ gives $x^2 + x-2≤0$. Hence we get $x≤1$ when $x≥1$.

ii) When $x<1$ we get $x^2 + |x-1| = x^2 -x +1$ and substituting into the inequality we get $x^2 - x +1 ≤ 1$ therfore $x^2-x ≤ 0$, factorising we get $x(x-1)≤0$, so we have $x≤0$ and $x≤1$

Not sure if this is correct, I feel there's a contradiction in the second part. I would really appreciate the help.

Answer 1

You have almost got it right but the final conclusion is as follows: for $x \geq 1$ we get $x=1$ and for $x <1$ the answer is $0\leq x <1$. Hence the final answer is $0\leq x \leq 1$.

Answer 2

You may rewrite the inequality first:

$$x^2 + |x-1| \leq 1 \Leftrightarrow |1-x| \leq 1-x^2 = (1-x)(1+x)$$

• $\color{blue}{x= 1}$ is an obvious solution.

For $x \neq 1$ you get

$$|1-x| \leq (1-x)(1+x) \stackrel{\color{blue}{x \neq 1}}{\Longleftrightarrow} \begin{cases} 1 \leq -(1+x) & \mbox{ for } x > 1 \\ 1\leq 1+x & \mbox{ for } x <1 \end{cases} \Leftrightarrow \begin{cases} x \leq -2 & \mbox{ for } x > 1 \color{red}{\mbox{ no solution}} \\ \color{blue}{0 \leq x} & \mbox{ for } \color{blue}{x <1} \end{cases}$$

All together $$\color{blue}{\boxed{0 \leq x \leq 1}}$$