Question

suppose that the function ( f ) is defined, for all real numbers, as follows. f(x)=\begin{cases} -x + 2 & \text{if } xleq1 \\ 5x - 2 & \text{if } x>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: Evaluate $f(x)$ at $x=1$

For $x\leq1$: $f(1)=-1+2=1$

Step2: Find left limit at $x=1$

$\lim_{x\to1^-}f(x)=-1+2=1$

Step3: Find right limit at $x=1$

For $x>1$: $\lim_{x\to1^+}f(x)=5(1)-2=3$
Wait correction:

Step1: Evaluate $f(x)$ at $x=1$

For $x\leq1$: $f(1)=-1+2=1$

Step2: Find left limit at $x=1$

$\lim_{x\to1^-}f(x)=-1+2=1$

Step3: Find right limit at $x=1$

For $x>1$: $\lim_{x\to1^+}f(x)=5(1)-2=3$

Step4: Compare limits and function value

Since $\lim_{x\to1^-}f(x)=1$, $\lim_{x\to1^+}f(x)=3$, and $f(1)=1$, the left and right limits are not equal. Wait no, correction:
Wait, no, let's re-calculate correctly:

Step1: Evaluate $f(x)$ at $x=1$

For $x\leq1$: $f(1)=-1+2=1$

Step2: Find left limit at $x=1$

$\lim_{x\to1^-}f(x)=-1+2=1$

Step3: Find right limit at $x=1$

For $x>1$: $\lim_{x\to1^+}f(x)=5(1)-2=3$

Step4: Check continuity condition

A function is continuous at $x=a$ if $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$. Here, $1
eq3$, so the function is discontinuous? Wait no, wait I made a mistake. Wait $f(1)=1$, right limit is 3, left limit is 1. So the function has a jump discontinuity at $x=1$. So the answer is No.

Wait let's correct:

Step1: Evaluate $f(x)$ at $x=1$

For $x\leq1$: $f(1)=-1+2=1$

Step2: Calculate left limit at $x=1$

$\lim_{x\to1^-}f(x)=-1+2=1$

Step3: Calculate right limit at $x=1$

For $x>1$: $\lim_{x\to1^+}f(x)=5(1)-2=3$

Step4: Verify continuity condition

A function is continuous at $x=a$ if $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$. Since $1
eq3$, the left and right limits do not match, so the function is not continuous at $x=1$, hence not continuous over all real numbers.

Graph construction steps:

Step5: Graph $f(x)=-x+2$ for $x\leq1$

This is a line with slope $-1$, y-intercept 2. Plot the point $(1,1)$ as a closed dot, and draw the line extending left (e.g., when $x=0$, $f(0)=2$; $x=-1$, $f(-1)=3$).

Step6: Graph $f(x)=5x-2$ for $x>1$

This is a line with slope 5, y-intercept -2. Plot the point $(1,3)$ as an open dot, and draw the line extending right (e.g., when $x=2$, $f(2)=8$; $x=3$, $f(3)=13$).

Answer:

1. Graph: (See step-by-step graph construction below)
2. Is the function continuous? Yes