Question

1-96. graph each piecewise - defined function given below and determine whether the function is continuous.
a. $f(x)=\begin{cases}x + 4&for xlt - 2\\frac{1}{2}x^{2}&for xgeq - 2end{cases}$
b. $f(x)=\begin{cases}x + 4&for xgeq3\\frac{1}{2}x^{2}&for xlt3end{cases}$

Explanation:

Step1: Analyze left - hand limit for part a

For \(x\to - 2^{-}\), \(f(x)=x + 4\). So \(\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{-}}(x + 4)=-2 + 4=2\).

Step2: Analyze right - hand limit for part a

For \(x\to - 2^{+}\), \(f(x)=\frac{1}{2}x^{2}\). So \(\lim_{x\to - 2^{+}}f(x)=\frac{1}{2}\times(-2)^{2}=2\). And \(f(-2)=\frac{1}{2}\times(-2)^{2}=2\). Since \(\lim_{x\to - 2^{-}}f(x)=\lim_{x\to - 2^{+}}f(x)=f(-2)\), the function is continuous at \(x=-2\).

Step3: Analyze left - hand limit for part b

For \(x\to3^{-}\), \(f(x)=\frac{1}{2}x^{2}\). So \(\lim_{x\to3^{-}}f(x)=\frac{1}{2}\times3^{2}=\frac{9}{2}\).

Step4: Analyze right - hand limit for part b

For \(x\to3^{+}\), \(f(x)=x + 4\). So \(\lim_{x\to3^{+}}f(x)=3 + 4=7\). Since \(\lim_{x\to3^{-}}f(x)
eq\lim_{x\to3^{+}}f(x)\), the function is discontinuous at \(x = 3\).

Answer:

a. The function is continuous.
b. The function is discontinuous.