a. Use limits to find the derivative function \( f^{\prime} \) for the function \( f \). b. Evaluate \( f^{\prime}(a) \) for the given values of \( a \). \( f(x)=\frac{4}{x+4} ; a=-\frac{1}{2}, 2 \) a. \( f^{\prime}(x)=\square \)

To find the derivative function \( f' \) for the function \( f(x) = \frac{4}{x+4} \) using limits, we will use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] ### Step 1: Calculate \( f(x+h) \) First, we need to find \( f(x+h) \): \[ f(x+h) = \frac{4}{(x+h) + 4} = \frac{4}{x + h + 4} \] ### Step 2: Set up the limit Now we can set up the limit for the derivative: \[ f'(x) = \lim_{h \to 0} \frac{\frac{4}{x+h+4} - \frac{4}{x+4}}{h} \] ### Step 3: Combine the fractions To combine the fractions in the numerator, we need a common denominator: \[ \frac{4}{x+h+4} - \frac{4}{x+4} = \frac{4(x+4) - 4(x+h+4)}{(x+h+4)(x+4)} \] Simplifying the numerator: \[ 4(x+4) - 4(x+h+4) = 4x + 16 - (4x + 4h + 16) = -4h \] So we have: \[ f'(x) = \lim_{h \to 0} \frac{-4h}{h \cdot (x+h+4)(x+4)} \] ### Step 4: Simplify the expression We can cancel \( h \) in the numerator and denominator (as long as \( h \neq 0 \)): \[ f'(x) = \lim_{h \to 0} \frac{-4}{(x+h+4)(x+4)} \] ### Step 5: Evaluate the limit Now we can evaluate the limit as \( h \to 0 \): \[ f'(x) = \frac{-4}{(x+4)(x+4)} = \frac{-4}{(x+4)^2} \] Thus, the derivative function is: \[ f'(x) = \frac{-4}{(x+4)^2} \] ### Step 6: Evaluate \( f'(a) \) for \( a = -\frac{1}{2} \) and \( a = 2 \) 1. For \( a = -\frac{1}{2} \): \[ f'\left(-\frac{1}{2}\right) = \frac{-4}{\left(-\frac{1}{2} + 4\right)^2} = \frac{-4}{\left(\frac{7}{2}\right)^2} = \frac{-4}{\frac{49}{4}} = \frac{-4 \cdot 4}{49} = \frac{-16}{49} \] 2. For \( a = 2 \): \[ f'(2) = \frac{-4}{(2 + 4)^2} = \frac{-4}{6^2} = \frac{-4}{36} = \frac{-1}{9} \] ### Final Answers a. The derivative function is: \[ f'(x) = \frac{-4}{(x+4)^2} \] b. The evaluated derivatives are: \[ f'\left(-\frac{1}{2}\right) = \frac{-16}{49}, \quad f'(2) = \frac{-1}{9} \]