Question

part: 0 / 3
(a) graph ( a(x) = -2x ) for ( x < 1 ).
(b) graph ( b(x) = sqrt{x - 1} ) for ( x geq 1 ).
(c) graph ( c(x) = \begin{cases} -2x & \text{for } x < 1 \\sqrt{x - 1} & \text{for } x geq 1 end{cases} )

Explanation:

Step1: Analyze $a(x) = -2x$ for $x<1$

This is a linear function with slope $-2$ and y-intercept $0$. We plot the line for all $x$ values less than 1, with an open circle at $x=1$ (since $x=1$ is not included):
At $x=1$, $a(1) = -2(1) = -2$, so open circle at $(1, -2)$.
Sample points: $x=0$, $a(0)=0$; $x=-1$, $a(-1)=2$.

Step2: Analyze $b(x) = \sqrt{x-1}$ for $x\geq1$

This is a square root function shifted right 1 unit. The domain is $x\geq1$, range is $y\geq0$. We plot for $x\geq1$, with a closed circle at $x=1$:
At $x=1$, $b(1) = \sqrt{1-1}=0$, so closed circle at $(1, 0)$.
Sample points: $x=2$, $b(2)=\sqrt{1}=1$; $x=5$, $b(5)=\sqrt{4}=2$.

Step3: Analyze piecewise $c(x)$

For $x<1$: $c(x)=-2x$, same as $a(x)$ (open circle at $(1, -2)$).
For $x\geq1$: $c(x)=\sqrt{x-1}$, same as $b(x)$ (closed circle at $(1, 0)$). Combine these two segments to form the piecewise graph.

Answer:

(a) For $a(x)=-2x$ ($x<1$): A downward-sloping line passing through $(0,0)$, $(-1,2)$, with an open circle at $(1,-2)$, extending left infinitely.
(b) For $b(x)=\sqrt{x-1}$ ($x\geq1$): A curve starting at closed circle $(1,0)$, passing through $(2,1)$ and $(5,2)$, increasing slowly as $x$ increases.
(c) For $c(x)$: The left segment is the line $y=-2x$ for $x<1$ (open circle at $(1,-2)$), and the right segment is the curve $y=\sqrt{x-1}$ for $x\geq1$ (closed circle at $(1,0)$), forming a piecewise graph with a break at $x=1$.