Question

1. $f(x)=\

$$\begin{cases}-x + 1&\\text{if }x < -2\\\\5&\\text{if }x\\geq -2\\end{cases}$$

$

1. $f(x)=\

$$\begin{cases}-\\frac{3}{2}x&\\text{if }x\\leq 0\\\\x + 1&\\text{if }0 < x\\leq 3\\end{cases}$$

$

1. $f(x)=\

$$\begin{cases}2x + 3&\\text{if }x < -1\\\\-|x|&\\text{if }-1\\leq x\\leq 1\\end{cases}$$

$
graph the following functions.

Explanation:

Problem 50:

Step1: Analyze first piece ($x < -2$)

The function is $f(x) = -x + 1$, a line with slope $-1$, y-intercept $1$. For $x=-2$, $f(-2) = -(-2)+1=3$, use an open circle at $(-2, 3)$ since $x < -2$. Plot points like $(-3, 4)$ and extend left.

Step2: Analyze second piece ($x \geq -2$)

The function is $f(x)=5$, a horizontal line. Use a closed circle at $(-2, 5)$ since $x \geq -2$, extend right indefinitely.

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Problem 51:

Step1: Analyze first piece ($x \leq 0$)

The function is $f(x)=-\frac{3}{2}x$, a line with slope $-\frac{3}{2}$, passes through origin. Use closed circle at $(0,0)$ since $x \leq 0$. Plot points like $(-2, 3)$ and extend left.

Step2: Analyze second piece ($0 < x \leq 3$)

The function is $f(x)=x+1$, a line with slope $1$, y-intercept $1$. Use open circle at $(0,1)$ and closed circle at $(3, 4)$ (since $x \leq 3$). Plot points like $(1,2)$, $(2,3)$ and connect to $(3,4)$.

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Problem 52:

Step1: Analyze first piece ($x < -1$)

The function is $f(x)=2x+3$, a line with slope $2$, y-intercept $3$. For $x=-1$, $f(-1)=2(-1)+3=1$, use open circle at $(-1,1)$. Plot points like $(-2, -1)$ and extend left.

Step2: Analyze second piece ($-1 \leq x \leq 1$)

The function is $f(x)=-|x|$, a V-shaped graph opening downward, vertex at $(0,0)$. Use closed circles at $(-1, -1)$ and $(1, -1)$ (since endpoints are included). Plot points $(-1,-1)$, $(0,0)$, $(1,-1)$ and connect.

Answer:

1. For $f(x)=

$$\begin{cases}-x + 1 & \text{if } x < -2 \\5 & \text{if } x \geq -2\end{cases}$$

$:

• Left segment: Line $y=-x+1$, open at $(-2,3)$, extending left.
• Right segment: Horizontal line $y=5$, closed at $(-2,5)$, extending right.

1. For $f(x)=

$$\begin{cases}-\frac{3}{2}x & \text{if } x \leq 0 \\x + 1 & \text{if } 0 < x \leq 3\end{cases}$$

$:

• Left segment: Line $y=-\frac{3}{2}x$, closed at $(0,0)$, extending left.
• Right segment: Line $y=x+1$, open at $(0,1)$, closed at $(3,4)$, connecting the two points.

1. For $f(x)=

$$\begin{cases}2x + 3 & \text{if } x < -1 \\-|x| & \text{if } -1 \leq x \leq 1\end{cases}$$

$:

• Left segment: Line $y=2x+3$, open at $(-1,1)$, extending left.
• Middle segment: Downward V-shape $y=-|x|$, closed at $(-1,-1)$ and $(1,-1)$, vertex at $(0,0)$.