Question

attempt 1: 10 attempts 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 none. answer: (\boxed{quad}) submit answer next item

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 \(\lim_{x\to - 3^{-}}(3x + 5)\). 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 \(\lim_{x\to - 3^{+}}(x^{2}+1)\). Substitute \(x = - 3\) into \(x^{2}+1\): \((-3)^{2}+1=9 + 1 = 10\).

Step3: Compare left - hand and right - hand limits

Since \(\lim_{x\to - 3^{-}}f(x)=-4\) and \(\lim_{x\to - 3^{+}}f(x)=10\), and \(-4
eq10\), the function has a discontinuity at \(x=-3\).

Answer:

\(x=-3\)