Question

1. given (f(x)=\begin{cases}-x^{2}+1, & \text{if }x < 2\\1 - 2x, & \text{if }xgeq2end{cases})

a) find (lim_{x
ightarrow2^{-}}f(x)=-(2)^{2}+1=-3), (lim_{x
ightarrow2^{+}}f(x)=1 - 2(2)=-3), (lim_{x
ightarrow2}f(x)=-3)
b) find (f(2)=1 - 2(2)=-3)
c) is (f(x)) continuous at (x = 2?) yes, because

Explanation:

Step1: Recall the definition of continuity

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here $a = 2$, $f(x)=

$$\begin{cases}-x^{2}+1, &x < 2\\1 - 2x, &x\geq2\end{cases}$$

$

Step2: Calculate the left - hand limit

For $x
ightarrow2^{-}$, we use the formula $f(x)=-x^{2}+1$. So $\lim_{x
ightarrow2^{-}}f(x)=-(2)^{2}+1=-4 + 1=-3$.

Step3: Calculate the right - hand limit

For $x
ightarrow2^{+}$, we use the formula $f(x)=1-2x$. So $\lim_{x
ightarrow2^{+}}f(x)=1-2\times2=1 - 4=-3$.

Step4: Calculate the function value at $x = 2$

When $x = 2$, we use the formula $f(x)=1-2x$. So $f(2)=1-2\times2=-3$.

Step5: Check the continuity condition

Since $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2)=-3$, the function $f(x)$ is continuous at $x = 2$.

Answer:

Yes, because $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2)=-3$