question 36 (1 point) discuss the continuity for the function. be sure to include: a)whether it is…

Explanation:

Step1: Find left - hand limit

We find $\lim_{x\rightarrow0^{-}}f(x)$. Since $x < 0$, $f(x)=4x + 2$. Then $\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}(4x + 2)=4\times0+2 = 2$.

Step2: Find right - hand limit

We find $\lim_{x\rightarrow0^{+}}f(x)$. Since $x\geq0$, $f(x)=x^{2}-1$. Then $\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow0^{+}}(x^{2}-1)=0^{2}-1=-1$.

Step3: Evaluate the function at $x = 0$

Since $x = 0$ and we use $f(x)=x^{2}-1$ (because $x = 0$ satisfies $x\geq0$), $f(0)=0^{2}-1=-1$.

Step4: Determine continuity

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)$. Here, $\lim_{x\rightarrow0^{-}}f(x)=2$, $\lim_{x\rightarrow0^{+}}f(x)=-1$ and $f(0)=-1$. Since $\lim_{x\rightarrow0^{-}}f(x)\neq\lim_{x\rightarrow0^{+}}f(x)$, the function is discontinuous at $x = 0$.

Step5: Identify the type of discontinuity

Since $\lim_{x\rightarrow0^{-}}f(x)$ and $\lim_{x\rightarrow0^{+}}f(x)$ both exist but are not equal, the type of discontinuity is a jump discontinuity.

Answer:

a) The function is discontinuous at $x = 0$ because $\lim_{x\rightarrow0^{-}}f(x)=2$ and $\lim_{x\rightarrow0^{+}}f(x)=-1$, and the left - hand limit is not equal to the right - hand limit. b) The type of discontinuity is a jump discontinuity.