91Ó°ÊÓ

Let the two functions be defined as follows: \(f(x) = \frac{1}{x+1}\), \(g(x) = \frac{x}{x-1}\).

Let's find \(f+g\). To do this, we simply add the two functions together: \((f+g)(x) = f(x) + g(x)\) \((f+g)(x) = \frac{1}{x+1} + \frac{x}{x-1}\).

Now we'll find \(f-g\). To do this, we subtract the second function from the first: \((f-g)(x) = f(x) - g(x)\) \((f-g)(x) = \frac{1}{x+1} - \frac{x}{x-1}\).

Next, we'll find the product \(fg\). To do this, we multiply the two functions together: \((fg)(x) = f(x) \cdot g(x)\) \((fg)(x) = \frac{1}{x+1} \cdot \frac{x}{x-1}\).

Finally, we'll find the quotient \(\frac{f}{g}\). To do this, we divide the first function by the second: \(\frac{f}{g}(x) = \frac{f(x)}{g(x)}\) \(\frac{f}{g}(x) = \frac{\frac{1}{x+1}}{\frac{x}{x-1}}\).

To find the domain of each function, we need to make sure the denominator is not zero. For each of the new functions, we will find the domain based on the domains of \(f(x)\) and \(g(x)\). (a) The domain of \(f(x)\) is \(x \neq -1\), and the domain of \(g(x)\) is \(x \neq 1\). Thus, the domain for \((f+g)(x)\) is \(x \neq -1, x \neq 1\). (b) Similarly, for \((f-g)(x)\), the domain is also \(x \neq -1, x \neq 1\). (c) For \((fg)(x)\), the domain is still \(x \neq -1, x \neq 1\). (d) To find the domain for \(\frac{f}{g}(x)\), we must make sure the denominator of \(g(x)\) is not zero. The domain for \(g(x)\) is already \(x \neq 1\), but since we are considering division, we should also consider the cases where \(g(x) = 0\). Plugging into the function, we get: \(g(x) = 0\) \(\frac{x}{x-1} = 0\) \(x = 0\) Thus, the domain for \(\frac{f}{g}(x)\) is \(x \neq -1, x \neq 0, x \neq 1\).