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Question
question 4 points 3 graph the piece-wise defined function: ( f(x) = \begin{cases} 3 + x, & \text{if } x < 1 \\ 2^x, & \text{if } x geq 1 end{cases} )
Step1: Analyze the first piece ($x < 1$)
The first piece is $f(x)=3 + x$ for $x < 1$. This is a linear function with slope $1$ and y - intercept $3$. To graph it, we can find some points. When $x = 0$, $f(0)=3+0 = 3$. When $x = 1$, $f(1)=3 + 1=4$, but since $x < 1$, the point $(1,4)$ is an open circle. We can draw a line with slope $1$ passing through $(0,3)$ and with an open circle at $(1,4)$.
Step2: Analyze the second piece ($x\geq1$)
The second piece is $f(x)=2^{x}$ for $x\geq1$. This is an exponential function. When $x = 1$, $f(1)=2^{1}=2$. When $x = 2$, $f(2)=2^{2}=4$. When $x = 0$, $f(0)=2^{0}=1$ (but we only consider $x\geq1$). We start with a closed circle at $(1,2)$ (since $x = 1$ is included here) and then plot the points for $x>1$ using the exponential growth property of $y = 2^{x}$.
Step3: Combine the two pieces
We draw the line $y = 3 + x$ for all $x < 1$ (with an open circle at $(1,4)$) and the exponential curve $y = 2^{x}$ for all $x\geq1$ (with a closed circle at $(1,2)$). We can see that at $x = 1$, the left - hand limit from the first function is $\lim_{x
ightarrow1^{-}}f(x)=3 + 1=4$ and the value from the second function is $f(1)=2$, so there is a jump discontinuity at $x = 1$.
(Note: Since the question is to graph the function, the above steps describe how to construct the graph. If we were to describe the graph in words, we have a linear segment for $x < 1$ and an exponential segment for $x\geq1$ with a jump at $x = 1$.)
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To graph $f(x)=
$:
- For $y = 3 + x$ ($x < 1$): Plot points like $(0,3)$, draw a line with slope $1$, mark an open circle at $(1,4)$.
- For $y = 2^{x}$ ($x\geq1$): Plot points like $(1,2)$, $(2,4)$, draw the exponential curve, mark a closed circle at $(1,2)$.
The graph has a linear part (left of $x = 1$) and an exponential part (right of and including $x = 1$) with a jump at $x = 1$.