Find the domain and range of the function f(x) = racx−1|x|.

The domain of a function is the set of all possible input values ($x$) for which the function is defined.

• The numerator $x-1$ is defined for all real numbers.
• The denominator $|x|$ cannot be zero, because division by zero is undefined. Set the denominator equal to zero to find the excluded value:

$$|x| = 0$$ $$x = 0$$

So, $x$ cannot be equal to 0. The domain is all real numbers except 0. In interval notation, this is $(-\infty, 0) \cup (0, \infty)$.

The range of a function is the set of all possible output values ($y$) that the function can produce.
Let $y = \frac{x-1}{|x|}$. We need to consider two cases for $x$:

Case 1: $x > 0$
If $x > 0$, then $|x| = x$.
The function becomes:

$$y = \frac{x-1}{x}$$ $$y = 1 - \frac{1}{x}$$

As $x$ takes values greater than 0:

• As $x \to 0^+$ (approaches 0 from the positive side), $\frac{1}{x} \to \infty$, so $y = 1 - \infty \to -\infty$.
• As $x \to \infty$, $\frac{1}{x} \to 0$, so $y = 1 - 0 \to 1$. For $x > 0$, the values of $y$ are in the interval $(-\infty, 1)$.

Case 2: $x < 0$
If $x < 0$, then $|x| = -x$.
The function becomes:

$$y = \frac{x-1}{-x}$$ $$y = -\left(\frac{x-1}{x}\right)$$ $$y = -\left(1 - \frac{1}{x}\right)$$ $$y = -1 + \frac{1}{x}$$

As $x$ takes values less than 0:

• As $x \to 0^-$ (approaches 0 from the negative side), $\frac{1}{x} \to -\infty$, so $y = -1 + (-\infty) \to -\infty$.
• As $x \to -\infty$, $\frac{1}{x} \to 0$, so $y = -1 + 0 \to -1$. For $x < 0$, the values of $y$ are in the interval $(-\infty, -1)$.

The total range is the union of the ranges from Case 1 and Case 2:
Range $= (-\infty, 1) \cup (-\infty, -1)$
The union of these two intervals is $(-\infty, 1)$.

The domain and range are:
Domain: $\boxed{x \in (-\infty, 0) \cup (0, \infty)}$
Range: $\boxed{y \in (-\infty, 1)}$