Question

for the following function, determine \\( \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \\).
\\( f(x) = \frac{6}{x + 9} \\)
\\( \frac{6}{(a + 9)^2} \\)
\\( -\frac{6}{a^2} \\)
\\( -\frac{6}{(a + 9)^2} \\)
\\( -\frac{6}{a + 9} \\)

Explanation:

Step1: Substitute \( f(a+h) \) and \( f(a) \)

Given \( f(x) = \frac{6}{x + 9} \), so \( f(a+h) = \frac{6}{(a + h) + 9} \) and \( f(a) = \frac{6}{a + 9} \). Substitute into the limit:

$$ \lim_{h \to 0} \frac{\frac{6}{(a + h) + 9} - \frac{6}{a + 9}}{h} $$

Step2: Simplify the numerator

Find a common denominator for the fractions in the numerator:

$$ \frac{6(a + 9) - 6((a + h) + 9)}{((a + h) + 9)(a + 9)} = \frac{6a + 54 - 6a - 6h - 54}{((a + h) + 9)(a + 9)} = \frac{-6h}{((a + h) + 9)(a + 9)} $$

Now the limit becomes:

$$ \lim_{h \to 0} \frac{\frac{-6h}{((a + h) + 9)(a + 9)}}{h} $$

Step3: Cancel \( h \) and take the limit

Cancel \( h \) (since \( h
eq 0 \) when taking the limit as \( h \to 0 \)):

$$ \lim_{h \to 0} \frac{-6}{((a + h) + 9)(a + 9)} $$

Now substitute \( h = 0 \):

$$ \frac{-6}{(a + 0 + 9)(a + 9)} = -\frac{6}{(a + 9)^2} $$

Answer:

\(-\dfrac{6}{(a + 9)^2}\) (corresponding to the option: \(-\dfrac{6}{(a + 9)^2}\))