Questions: Given f(x)=1/(x+7), find the average rate of change of f(x) on the interval [2,2+h]. Your answer will be an expression involving h.

Solution Steps

We are given the function \( f(x) = \frac{1}{x+7} \) and need to find the average rate of change of this function over the interval \([2, 2+h]\).

The average rate of change of a function \( f \) over the interval \([x_1, x_2]\) is defined as:

\[ \text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]

In our case, \( x_1 = 2 \) and \( x_2 = 2 + h \). Thus, we need to evaluate:

\[ f(2) = \frac{1}{2 + 7} = \frac{1}{9} \]

\[ f(2+h) = \frac{1}{(2+h) + 7} = \frac{1}{9 + h} \]

Now, substituting these values into the average rate of change formula:

\[ \text{Average Rate of Change} = \frac{\frac{1}{9+h} - \frac{1}{9}}{(2+h) - 2} = \frac{\frac{1}{9+h} - \frac{1}{9}}{h} \]

To simplify the expression, we find a common denominator for the fractions in the numerator:

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

Substituting this back into the average rate of change formula gives:

\[ \text{Average Rate of Change} = \frac{\frac{-h}{(9+h) \cdot 9}}{h} = \frac{-1}{(9+h) \cdot 9} \]

Final Answer