<s>[INST] Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$ 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.
 [/INST] 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
$$\color{blue}{\boxed{0 \leq x \leq 1}}$$
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