Questions: Question 24 let (f(x)=frac1x+9). (a) Determine the difference quotient and simplify. [ fracf(x+h)-f(x)h= ] (b) Determine (f^prime(x)) by letting (h rightarrow 0) in part (a). [ f^prime(x)= ]

Solution Steps

(a) To determine the difference quotient for the function \( f(x) = \frac{1}{x+9} \), we need to calculate \( \frac{f(x+h) - f(x)}{h} \). This involves substituting \( f(x+h) = \frac{1}{x+h+9} \) into the expression and simplifying the resulting fraction.

(b) To find \( f'(x) \), we take the limit of the difference quotient as \( h \rightarrow 0 \). This involves applying the limit to the simplified expression from part (a).

We start with the function \( f(x) = \frac{1}{x + 9} \). The difference quotient is given by:

\[ \frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{x+h+9} - \frac{1}{x+9}}{h} \]

To simplify this expression, we find a common denominator:

\[ = \frac{(x + 9) - (x + h + 9)}{h \cdot (x + h + 9)(x + 9)} = \frac{-h}{h \cdot (x + h + 9)(x + 9)} \]

This simplifies to:

\[ = -\frac{1}{(x + 9)(x + h + 9)} \]

Next, we find the derivative \( f'(x) \) by taking the limit of the difference quotient as \( h \) approaches 0:

\[ f'(x) = \lim_{h \to 0} -\frac{1}{(x + 9)(x + h + 9)} = -\frac{1}{(x + 9)(x + 9)} = -\frac{1}{(x + 9)^2} \]

Final Answer