Show that \( f(x)=\frac{2}{x+1}+1 \)

To show that \( f(x)=\frac{2}{x+1}+1 \) is well-defined and behaves as expected, we can analyze its components and consider its properties. First, let's rewrite the function: \[ f(x) = \frac{2}{x+1} + 1 \] We can find the domain of this function. The only restriction comes from the denominator \( x + 1 \) not being equal to zero. Therefore, \( x \neq -1 \). The function is defined for all real numbers except \( -1 \). Next, we can simplify the function further: \[ f(x) = \frac{2}{x+1} + \frac{x + 1}{x + 1} = \frac{2 + (x + 1)}{x + 1} = \frac{x + 3}{x + 1} \] So now \( f(x) = \frac{x + 3}{x + 1} \). This shows that the function is indeed a rational function. We can also explore the behavior of the function as \( x \) approaches \( -1 \). As \( x \) approaches \( -1 \) from either side, \( f(x) \) tends toward \( \pm \infty\) indicating a vertical asymptote at \( x = -1 \). To evaluate the limits at infinity: - As \( x \to \infty \), \( f(x) \to 1 \) - As \( x \to -\infty \), \( f(x) \to 1 \) These results indicate that the horizontal asymptote of the function is \( y = 1 \). In conclusion, \( f(x) = \frac{2}{x+1} + 1 \) is a well-defined function on \( \mathbb{R} \setminus \{-1\}\), and it has interesting asymptotic behavior worth noting!