Question

for the following function, \\( f(x) = \frac{6}{x + 9} \\) (options: \\( \frac{6}{(a + 9)^2} \\), \\( -\frac{6}{a^2} \\), \\( -\frac{6}{(a + 9)^2} \\), \\( -\frac{6}{a + 9} \\))
5 multiple choice 1 point
is \\( f(x) \\) continuous at \\( x = 4 \\)?
\\( f(x) = \

$$\begin{cases} 3x + 3 & \\text{if } x \\leq -4 \\\\ -x^2 & \\text{if } -4 < x \\leq 4 \\\\ \\frac{2}{x} & \\text{if } x > 4 \\end{cases}$$

\\)
options: yes, no

Explanation:

Step1: Find \( f(4) \)

For \( x = 4 \), we use the piece \( -x^2 \) (since \( -4 < 4 \leq 4 \)). So \( f(4)=-(4)^2=-16 \).

Step2: Find left - hand limit as \( x\to4^- \)

As \( x \) approaches 4 from the left (\( x\to4^- \)), we still use the piece \( -x^2 \) (because \( x\leq4 \) for this piece). So \( \lim_{x\to4^-}f(x)=\lim_{x\to4^-}(-x^2)=-(4)^2 = - 16 \).

Step3: Find right - hand limit as \( x\to4^+ \)

As \( x \) approaches 4 from the right (\( x\to4^+ \)), we use the piece \( \frac{2}{x} \) (since \( x > 4 \) for this piece). So \( \lim_{x\to4^+}f(x)=\lim_{x\to4^+}\frac{2}{x}=\frac{2}{4}=\frac{1}{2} \).

Step4: Check continuity conditions

For a function to be continuous at \( x = a \), we need \( \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a) \). Here, \( \lim_{x\to4^-}f(x)=-16 \), \( \lim_{x\to4^+}f(x)=\frac{1}{2} \), and \( f(4)=-16 \). Since \( \lim_{x\to4^-}f(x)
eq\lim_{x\to4^+}f(x) \), the function is not continuous at \( x = 4 \).

Answer:

No (the option "No")