Question

suppose that the function f is defined, for all real numbers, as follows.
$f(x)=\

$$\begin{cases} 2x - 1 & \\text{if } x < 1 \\\\ -x + 2 & \\text{if } x \\geq 1 \\end{cases}$$

$
graph the function f. then determine whether or not the function is continuous.
is the function continuous?
\bigcirc yes
\bigcirc no

Explanation:

Step1: Analyze the left piece ($x < 1$)

The function is $f(x) = 2x - 1$ for $x < 1$. This is a linear function with slope $2$ and y - intercept $-1$. Let's find the value as $x$ approaches $1$ from the left. Using the limit as $x \to 1^-$, we substitute $x = 1$ into $2x - 1$: $2(1)-1=1$. Also, for the left - hand side, when $x < 1$, we can consider a test point, say $x = 0$, then $f(0)=2(0)-1=-1$.

Step2: Analyze the right piece ($x\geq1$)

The function is $f(x)=-x + 2$ for $x\geq1$. This is a linear function with slope $-1$ and y - intercept $2$. Let's find the value at $x = 1$: $f(1)=-1 + 2 = 1$. Also, let's find the limit as $x\to1^+$, substituting $x = 1$ into $-x + 2$ gives $-1+2 = 1$.

Step3: Check continuity conditions

For a function to be continuous at a point $a$, three conditions must be met:

1. $f(a)$ is defined. Here, at $a = 1$, $f(1)=1$ (from the right - hand piece).
2. $\lim_{x\to a^-}f(x)$ exists. $\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(2x - 1)=1$.
3. $\lim_{x\to a^+}f(x)$ exists. $\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(-x + 2)=1$.

And $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1) = 1$. So the function is continuous at $x = 1$. Since the function is composed of two linear functions (which are continuous on their respective domains) and is continuous at the point where the pieces meet ($x = 1$), the function is continuous everywhere.

Answer:

Yes (the option "Yes" for the question "Is the function continuous?")