Question

let
$f(x) = \

$$\begin{cases} 3x + 5 & \\text{if } x < -3 \\\\ x^2 + 1 & \\text{if } x \\geq -3 \\end{cases}$$

$
at what $x$ value is the function $f(x)$ discontinuous? if there is no discontinuity, enter
one\.
answer:
\boxed{}
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Explanation:

Step1: Find left - hand limit at \(x = - 3\)

For \(x\lt - 3\), \(f(x)=3x + 5\). The left - hand limit as \(x\to - 3^{-}\) is \(\lim_{x\to - 3^{-}}f(x)=\lim_{x\to - 3^{-}}(3x + 5)\). Substitute \(x=-3\) into \(3x + 5\): \(3\times(-3)+5=-9 + 5=-4\).

Step2: Find right - hand limit at \(x = - 3\)

For \(x\geq - 3\), \(f(x)=x^{2}+1\). The right - hand limit as \(x\to - 3^{+}\) is \(\lim_{x\to - 3^{+}}f(x)=\lim_{x\to - 3^{+}}(x^{2}+1)\). Substitute \(x = - 3\) into \(x^{2}+1\): \((-3)^{2}+1=9 + 1 = 10\).

Step3: Check the function value at \(x=-3\)

For \(x=-3\), since \(x\geq - 3\), \(f(-3)=(-3)^{2}+1 = 10\).
Since \(\lim_{x\to - 3^{-}}f(x)=-4\) and \(\lim_{x\to - 3^{+}}f(x)=10\), and \(\lim_{x\to - 3^{-}}f(x)
eq\lim_{x\to - 3^{+}}f(x)\), the function is discontinuous at \(x=-3\).

Answer:

\(x=-3\)