<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 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 $0\leq x \leq 1$.
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