Question

11 which function has a point of discontinuity at x = 3 when graphed? a $f(x) = \

$$\begin{cases} 3x + 1 & \\text{for } x < 3 \\\\ x^2 + 1 & \\text{for } x \\geq 3 \\end{cases}$$

$ c $f(x) = \frac{x - 3}{x^2}$ b $f(x) = |x - 3| + 2$ d $f(x) = \frac{x + 2}{x^2 - 9}$ \bigcirc a \bigcirc d \bigcirc b \bigcirc c

Explanation:

Step1: Analyze Option A

For the piece - wise function \(f(x)=

$$\begin{cases}3x + 1&x\lt3\\x^{2}+1&x\geq3\end{cases}$$

\), we find the left - hand limit as \(x\to3^{-}\): \(\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}(3x + 1)=3\times3+1 = 10\). The right - hand limit as \(x\to3^{+}\): \(\lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}(x^{2}+1)=3^{2}+1 = 10\). And \(f(3)=3^{2}+1 = 10\). Since \(\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{+}}f(x)=f(3)\), the function is continuous at \(x = 3\).

Step2: Analyze Option B

The function \(f(x)=\vert x - 3\vert+2\) is an absolute - value function. The absolute - value function \(y=\vert x - a\vert+b\) is continuous everywhere. For \(f(x)=\vert x - 3\vert+2\), we can rewrite it as \(f(x)=

$$\begin{cases}-(x - 3)+2&x\lt3\\x - 3 + 2&x\geq3\end{cases}$$

=

$$\begin{cases}-x+5&x\lt3\\x - 1&x\geq3\end{cases}$$

\). The left - hand limit as \(x\to3^{-}\): \(\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}(-x + 5)=-3 + 5 = 2\). The right - hand limit as \(x\to3^{+}\): \(\lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}(x - 1)=3 - 1 = 2\). And \(f(3)=\vert3 - 3\vert+2 = 2\). So the function is continuous at \(x = 3\).

Step3: Analyze Option C

The function \(f(x)=\frac{x - 3}{x^{2}}\). The denominator \(x^{2}\) is equal to \(0\) when \(x = 0\), not when \(x = 3\) (since when \(x = 3\), the denominator \(x^{2}=9
eq0\)). So the function is defined at \(x = 3\) and \(\lim_{x\to3}\frac{x - 3}{x^{2}}=\frac{3 - 3}{9}=0\), and \(f(3)=\frac{3 - 3}{9}=0\). The function is continuous at \(x = 3\).

Step4: Analyze Option D

The function \(f(x)=\frac{x + 2}{x^{2}-9}\). We factor the denominator: \(x^{2}-9=(x - 3)(x + 3)\). The denominator is equal to \(0\) when \(x=3\) or \(x=-3\). So the function is not defined at \(x = 3\). To check for discontinuity, we know that if a function is not defined at a point, it has a discontinuity at that point. Also, we can check the limit as \(x\to3\): \(\lim_{x\to3}\frac{x + 2}{(x - 3)(x + 3)}\). As \(x\to3\), the numerator approaches \(3 + 2=5\) and the denominator approaches \(0\) (since \(x-3\to0\) and \(x + 3\to6\)). So the limit does not exist (it goes to \(\pm\infty\)) and the function is not defined at \(x = 3\), so there is a point of discontinuity at \(x = 3\).

Answer:

D. \(f(x)=\frac{x + 2}{x^{2}-9}\)