Find the difference quotient \( \frac{f(x+h)-f(x)}{h} \), where \( h \neq 0 \), for the function below. \( f(x)=\frac{7}{x+1} \) Simplify. your answer as much as possible.

To find the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x)=\frac{7}{x+1} \), we start by calculating \( f(x+h) \): \[ f(x+h) = \frac{7}{(x+h)+1} = \frac{7}{x+h+1} \] Now, substituting \( f(x) \) and \( f(x+h) \) into the difference quotient, we have: \[ \frac{f(x+h)-f(x)}{h} = \frac{\frac{7}{x+h+1} - \frac{7}{x+1}}{h} \] To simplify this, we find a common denominator for the two fractions in the numerator: \[ \frac{7}{x+h+1} - \frac{7}{x+1} = \frac{7(x+1) - 7(x+h+1)}{(x+h+1)(x+1)} \] This simplifies to: \[ \frac{7(x+1 - (x+h+1))}{(x+h+1)(x+1)} = \frac{7(x + 1 - x - h - 1)}{(x+h+1)(x+1)} = \frac{7(-h)}{(x+h+1)(x+1)} \] Now substitute this back into the difference quotient: \[ \frac{\frac{7(-h)}{(x+h+1)(x+1)}}{h} = \frac{7(-h)}{h(x+h+1)(x+1)} \] Since \( h \neq 0 \), we can cancel \( h \): \[ = \frac{7(-1)}{(x+h+1)(x+1)} = \frac{-7}{(x+h+1)(x+1)} \] Thus, the simplified difference quotient is: \[ \frac{-7}{(x+h+1)(x+1)} \]