We need to solve the inequality $|x| > x^2$. Since $|x|$ is always non-negative, we can start by assuming that $x$ is also non-negative. In this case, the inequality simplifies to $x > x^2$. Dividing both sides by $x$ (which is positive) gives $1 > x$, which means $0 \leq x < 1$.
Now, let's consider the case where $x$ is negative. In this case, we can rewrite the inequality as $|x| > (-x)^2$, which simplifies to $|x| > x^2$. We can solve this inequality in the same way as before, giving $-1 < x \leq 0$.
Combining the two cases, we get the solution $x \in (-1, 0] \cup [0, 1)$.
Let $f(x) = |x|$ and $g(x) = x^2$.
Find all values of $x$ for which $f(x) > g(x)$.
2 answers
almost: (-1,0)U(0,1)
note: (-1,0]U[0,1) is just (-1,1)
note: (-1,0]U[0,1) is just (-1,1)