Question

f(x) = x - \sqrt{x^2 - 1}

Answer 1

• Domain: $(-\infty, -1] \cup [1, \infty)$

• Range: $(-\infty, -1] \cup (0, 1]$

• Inverse function: $f^{-1}(x) = \frac{x^2 + 1}{2x}$ for $x \in (-\infty, -1] \cup (0, 1]$.

Answer 2

The function given is $f(x) = x - \sqrt{x^2 - 1}$.

1. Domain: The term $\sqrt{x^2 - 1}$ requires $x^2 - 1 \ge 0$, which means $(x-1)(x+1) \ge 0$. This inequality holds for $x \le -1$ or $x \ge 1$. The domain of $f(x)$ is $D = (-\infty, -1] \cup [1, \infty)$.

2. Range:

   • Consider $x \ge 1$. We can rewrite $f(x)$ by multiplying by the conjugate: $f(x) = (x - \sqrt{x^2 - 1}) \frac{x + \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}} = \frac{x^2 - (x^2 - 1)}{x + \sqrt{x^2 - 1}} = \frac{1}{x + \sqrt{x^2 - 1}}$. For $x \ge 1$, $x$ is positive and $\sqrt{x^2 - 1} \ge 0$. Thus, $x + \sqrt{x^2 - 1}$ is positive. As $x \to \infty$, $x + \sqrt{x^2 - 1} \to \infty$, so $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{x + \sqrt{x^2 - 1}} = 0$. At $x = 1$, $f(1) = 1 - \sqrt{1^2 - 1} = 1$. The derivative $f'(x) = 1 - \frac{x}{\sqrt{x^2 - 1}}$ for $x > 1$. Since $\frac{x}{\sqrt{x^2 - 1}} = \sqrt{\frac{x^2}{x^2-1}} = \sqrt{1 + \frac{1}{x^2-1}} > 1$ for $x > 1$, $f'(x) < 0$. The function is decreasing on $[1, \infty)$. The range for $x \in [1, \infty)$ is $(0, 1]$.

   • Consider $x \le -1$. Let $x = -y$ where $y \ge 1$. $f(x) = f(-y) = -y - \sqrt{(-y)^2 - 1} = -y - \sqrt{y^2 - 1}$. As $y \to \infty$ (i.e., $x \to -\infty$), $-y \to -\infty$ and $-\sqrt{y^2 - 1} \to -\infty$. So, $\lim_{x \to -\infty} f(x) = -\infty$. At $x = -1$, $f(-1) = -1 - \sqrt{(-1)^2 - 1} = -1$. The derivative $f'(x) = 1 - \frac{x}{\sqrt{x^2 - 1}}$ for $x < -1$. Let $x = -y$ with $y > 1$. $f'(x) = 1 - \frac{-y}{\sqrt{(-y)^2 - 1}} = 1 + \frac{y}{\sqrt{y^2 - 1}}$. For $y > 1$, $\frac{y}{\sqrt{y^2 - 1}} > 0$, so $f'(x) > 1 > 0$. The function is increasing on $(-\infty, -1]$. The range for $x \in (-\infty, -1]$ is $(-\infty, -1]$.

   The total range of $f(x)$ is the union of the ranges from the two intervals: $R = (-\infty, -1] \cup (0, 1]$.

3. Injectivity: The function is strictly increasing on $(-\infty, -1]$ and strictly decreasing on $[1, \infty)$. Also, the ranges for these two intervals, $(-\infty, -1]$ and $(0, 1]$, are disjoint. This means that any value in the range is attained for a unique value of $x$ in the domain. Therefore, the function $f(x)$ is injective (one-to-one) on its domain.

4. Inverse Function: Since $f(x)$ is injective, its inverse $f^{-1}(x)$ exists. Let $y = x - \sqrt{x^2 - 1}$. We want to solve for $x$ in terms of $y$. $y - x = -\sqrt{x^2 - 1}$ $x - y = \sqrt{x^2 - 1}$ Squaring both sides: $(x - y)^2 = x^2 - 1$ $x^2 - 2xy + y^2 = x^2 - 1$ $-2xy + y^2 = -1$ $2xy = y^2 + 1$ $x = \frac{y^2 + 1}{2y}$. Thus, the inverse function is $f^{-1}(y) = \frac{y^2 + 1}{2y}$. Replacing $y$ with $x$, we get $f^{-1}(x) = \frac{x^2 + 1}{2x}$. The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $(-\infty, -1] \cup (0, 1]$. The range of $f^{-1}(x)$ is the domain of $f(x)$, which is $(-\infty, -1] \cup [1, \infty)$.

Summary of properties:

• Domain: $(-\infty, -1] \cup [1, \infty)$
• Range: $(-\infty, -1] \cup (0, 1]$
• Monotonicity: Increasing on $(-\infty, -1]$, Decreasing on $[1, \infty)$
• Injectivity: Injective
• Inverse function: $f^{-1}(x) = \frac{x^2 + 1}{2x}$ with domain $(-\infty, -1] \cup (0, 1]$.