c) \( y=\tan ^{-1} \frac{x}{1+x} \)

To analyze the function \( y = \tan^{-1} \left( \frac{x}{1+x} \right) \), we can explore its properties, such as its domain, range, and behavior. ### Domain The expression \( \frac{x}{1+x} \) is defined for all \( x \) except when the denominator is zero. The denominator \( 1+x = 0 \) when \( x = -1 \). Therefore, the domain of the function is: \[ \text{Domain: } x \in (-\infty, -1) \cup (-1, \infty) \] ### Range The function \( \tan^{-1}(u) \) has a range of \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) for any real number \( u \). Since \( \frac{x}{1+x} \) can take any real value as \( x \) approaches \( -1 \) from the left or right, the range of \( y \) will also be: \[ \text{Range: } y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] ### Behavior To understand the behavior of the function, we can analyze the limits as \( x \) approaches the critical point \( -1 \) and as \( x \) approaches \( \pm \infty \). 1. **As \( x \to -1^- \)**: \[ \frac{x}{1+x} \to \frac{-1}{0^-} \to +\infty \quad \Rightarrow \quad y \to \tan^{-1}(+\infty) = \frac{\pi}{2} \] 2. **As \( x \to -1^+ \)**: \[ \frac{x}{1+x} \to \frac{-1}{0^+} \to -\infty \quad \Rightarrow \quad y \to \tan^{-1}(-\infty) = -\frac{\pi}{2} \] 3. **As \( x \to -\infty \)**: \[ \frac{x}{1+x} \to 1 \quad \Rightarrow \quad y \to \tan^{-1}(1) = \frac{\pi}{4} \] 4. **As \( x \to +\infty \)**: \[ \frac{x}{1+x} \to 1 \quad \Rightarrow \quad y \to \tan^{-1}(1) = \frac{\pi}{4} \] ### Summary The function \( y = \tan^{-1} \left( \frac{x}{1+x} \right) \) has the following properties: - **Domain**: \( x \in (-\infty, -1) \cup (-1, \infty) \) - **Range**: \( y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) - **Behavior**: - Approaches \( \frac{\pi}{2} \) as \( x \to -1^- \) - Approaches \( -\frac{\pi}{2} \) as \( x \to -1^+ \) - Approaches \( \frac{\pi}{4} \) as \( x \to \pm \infty \) This function is continuous on its domain and has a vertical asymptote at \( x = -1 \).