Question

(10) 10. find the discontinuities and classify them as removable, jump or infinite
f(x)=\begin{cases}1&\text{if }x < - 2\\x + 1&\text{if }-2leq xleq0\\1/x&\text{if }0 < xleq1\\x^{2}&\text{if }x>1end{cases}
sketch the graph of (y = f(x))

Explanation:

Step1: Check continuity at $x = - 2$

Left - hand limit: $\lim_{x
ightarrow - 2^{-}}f(x)=1$. Right - hand limit: $\lim_{x
ightarrow - 2^{+}}f(x)=-2 + 1=-1$. Since $\lim_{x
ightarrow - 2^{-}}f(x)
eq\lim_{x
ightarrow - 2^{+}}f(x)$, there is a jump discontinuity at $x=-2$.

Step2: Check continuity at $x = 0$

Left - hand limit: $\lim_{x
ightarrow0^{-}}f(x)=0 + 1 = 1$. Right - hand limit: $\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}\frac{1}{x}=\infty$. So, there is an infinite discontinuity at $x = 0$.

Step3: Check continuity at $x = 1$

Left - hand limit: $\lim_{x
ightarrow1^{-}}f(x)=\frac{1}{1}=1$. Right - hand limit: $\lim_{x
ightarrow1^{+}}f(x)=1^{2}=1$. Also, $f(1) = 1$. The function is continuous at $x = 1$.

Answer:

There is a jump discontinuity at $x=-2$, an infinite discontinuity at $x = 0$, and the function is continuous at $x = 1$.