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

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 \(a = 3\) and \(b = 2\), \(f(x)\) is continuous at \(x = 3\) (in fact, continuous for all real \(x\)).

Step3: Analyze Option C

The function \(f(x)=\frac{x - 3}{x^{2}}\). The denominator \(x^{2}=0\) when \(x = 0\). To check continuity at \(x = 3\), we find \(f(3)=\frac{3 - 3}{3^{2}}=\frac{0}{9}=0\), and the limit as \(x\to3\) is \(\lim_{x\to3}\frac{x - 3}{x^{2}}=\frac{3 - 3}{3^{2}} = 0\). Also, the function is defined at \(x = 3\) (denominator is non - zero at \(x = 3\)), so it 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 zero when \(x=3\) or \(x=-3\). So the function is not defined at \(x = 3\), which means there is a point of discontinuity at \(x = 3\).

Answer:

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