Question

1. let (f(x)=\begin{cases}x - 2& \text{if }x>-1\\4 - x^{2}&\text{if }xleq - 1end{cases}). use the definition of continuity to show at (x = - 1), state the type of discontinuity if not

Explanation:

Step1: Recall continuity condition

A function $f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here $a=-1$, for $x < - 1$, $f(x)=4 - x^{2}$, and for $x>-1$, $f(x)=x - 2$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{-}}(4 - x^{2})$. Substitute $x=-1$ into $4 - x^{2}$, we get $4-(-1)^{2}=4 - 1=3$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow - 1^{+}}f(x)=\lim_{x
ightarrow - 1^{+}}(x - 2)$. Substitute $x = - 1$ into $x - 2$, we get $-1-2=-3$.

Step4: Analyze continuity

Since $\lim_{x
ightarrow - 1^{-}}f(x)=3$ and $\lim_{x
ightarrow - 1^{+}}f(x)=-3$, $\lim_{x
ightarrow - 1^{-}}f(x)
eq\lim_{x
ightarrow - 1^{+}}f(x)$. So the function is discontinuous at $x=-1$.

Step5: Determine type of discontinuity

Since the left - hand and right - hand limits exist but are not equal, the type of discontinuity is a jump discontinuity.

Answer:

The function $f(x)$ is discontinuous at $x = - 1$ and the type of discontinuity is a jump discontinuity.