Question

Solve for x: -1 ≤ x^2+1/2x ≤ 1

Answer 1

x ∈ {-1, 1}

Answer 2

The problem asks us to solve the inequality $-1 \leq x^2 + \frac{1}{2x} \leq 1$.

There is an ambiguity in the expression x^2+1/2x. It can be interpreted as $x^2 + \frac{1}{2x}$ or $\frac{x^2+1}{2x}$. Given the provided "similar question" which is $-1 \leq \frac{1+x^2}{2x} \leq 1$, it is highly probable that the original question intended to be $\frac{x^2+1}{2x}$. This is a common typographical issue where the fraction bar doesn't extend as intended. We will proceed with the assumption that the question is $-1 \leq \frac{x^2+1}{2x} \leq 1$.

The compound inequality $-1 \leq \frac{x^2+1}{2x} \leq 1$ can be rewritten in terms of absolute value as: $\left| \frac{x^2+1}{2x} \right| \leq 1$ Since $x^2+1$ is always non-negative for any real $x$, we can write $x^2+1 = |x^2+1|$. Also, $|2x| = 2|x|$. So the inequality becomes: $\frac{|x^2+1|}{|2x|} \leq 1$ $\frac{x^2+1}{2|x|} \leq 1$ Note that $x \neq 0$ because $2x$ is in the denominator. Since $2|x|$ is always positive for $x \neq 0$, we can multiply both sides by $2|x|$ without changing the direction of the inequality: $x^2+1 \leq 2|x|$ Rearrange the terms to one side: $x^2 - 2|x| + 1 \leq 0$ Let $y = |x|$. Since $x$ is a real number, $y \geq 0$. Substituting $y$ into the inequality: $y^2 - 2y + 1 \leq 0$ This expression is a perfect square trinomial: $(y-1)^2 \leq 0$ The square of any real number is always non-negative. That is, $(y-1)^2 \geq 0$. For $(y-1)^2$ to be both less than or equal to zero AND greater than or equal to zero, the only possibility is that it must be exactly zero: $(y-1)^2 = 0$ Taking the square root of both sides: $y-1 = 0$ $y = 1$ Now, substitute back $y = |x|$: $|x| = 1$ This implies that $x$ can be $1$ or $-1$. $x = 1 \quad \text{or} \quad x = -1$ Both these values satisfy the condition $x \neq 0$.

Let's verify these solutions in the original inequality: For $x=1$: $-1 \leq \frac{1^2+1}{2(1)} \leq 1$ $-1 \leq \frac{1+1}{2} \leq 1$ $-1 \leq \frac{2}{2} \leq 1$ $-1 \leq 1 \leq 1$ (This is true)

For $x=-1$: $-1 \leq \frac{(-1)^2+1}{2(-1)} \leq 1$ $-1 \leq \frac{1+1}{-2} \leq 1$ $-1 \leq \frac{2}{-2} \leq 1$ $-1 \leq -1 \leq 1$ (This is true)

Thus, the solutions are $x=1$ and $x=-1$.

The final answer is $\boxed{{x \in \{-1, 1\}}}$.

Explanation of the solution: The inequality $-1 \leq \frac{x^2+1}{2x} \leq 1$ is equivalent to $\left| \frac{x^2+1}{2x} \right| \leq 1$. Since $x^2+1 \geq 0$, this simplifies to $\frac{x^2+1}{2|x|} \leq 1$. Multiplying by $2|x|$ (which is positive for $x \neq 0$), we get $x^2+1 \leq 2|x|$. Rearranging gives $x^2 - 2|x| + 1 \leq 0$. Let $y=|x|$, so $y^2 - 2y + 1 \leq 0$, which is $(y-1)^2 \leq 0$. Since a square must be non-negative, the only possibility is $(y-1)^2 = 0$, which means $y=1$. Substituting back $|x|=y$, we get $|x|=1$, which implies $x=1$ or $x=-1$. These values satisfy the condition $x \neq 0$.

Answer: $x \in \{-1, 1\}$