Questions: Let f(x) = 1 / (x + 3). By taking the limit of the appropriate difference quotient, show that f'(-7) = -1 / 16.

Transcript text: Let $f(x)=\frac{1}{x+3}$. By taking the limit of the appropriate difference quotient, show that $f^{\prime}(-7)=\frac{-1}{16}$.

Solution

Solution Steps

To find the derivative of the function $ f(x) = \frac{1}{x+3} $ at $ x = -7 $, we use the definition of the derivative as a limit of the difference quotient. The difference quotient is given by $ \frac{f(x+h) - f(x)}{h} $. We will substitute $ f(x) $ into this expression and take the limit as $ h $ approaches 0 to find $ f'(-7) $.

Step 1: Define the Function and Difference Quotient

We start with the function $ f(x) = \frac{1}{x+3} $. To find the derivative at a specific point, we use the difference quotient:

\[ \frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{x+h+3} - \frac{1}{x+3}}{h} \]

Step 2: Simplify the Difference Quotient

Simplify the expression:

\[ \frac{\frac{1}{x+h+3} - \frac{1}{x+3}}{h} = \frac{\frac{x+3 - (x+h+3)}{(x+3)(x+h+3)}}{h} = \frac{-h}{h(x+3)(x+h+3)} \]

Cancel $ h $ in the numerator and denominator:

\[ = \frac{-1}{(x+3)(x+h+3)} \]

Step 3: Take the Limit as $ h \to 0 $

To find the derivative $ f'(x) $, take the limit of the simplified difference quotient as $ h $ approaches 0:

\[ f'(x) = \lim_{h \to 0} \frac{-1}{(x+3)(x+h+3)} = \frac{-1}{(x+3)^2} \]

Step 4: Evaluate the Derivative at $ x = -7 $