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Drag the tiles to the correct locations on the equation. Not all pieces will be used.

Consider this absolute value function:

[tex]\[ f(x) = |x + 3| \][/tex]

How can function [tex]\( f \)[/tex] be rewritten as a piecewise function?

[tex]\[
\begin{array}{c}
x \geq -3 \quad x-3 \quad -x-3 \quad x \geq 3 \quad x \ \textless \ 3 \quad x \ \textless \ -3 \quad -x+3 \\
f(x) = \left\{
\begin{array}{ll}
x + 3, & \text{if } x \geq -3 \\
-(x + 3), & \text{if } x \ \textless \ -3
\end{array}
\right.
\end{array}
\][/tex]


Sagot :

To rewrite the absolute value function [tex]\( f(x) = |x + 3| \)[/tex] as a piecewise function, we need to consider the definition of the absolute value. The absolute value function [tex]\( |x + 3| \)[/tex] can be rewritten depending on whether the expression inside the absolute value, [tex]\( x + 3 \)[/tex], is non-negative (i.e., zero or positive) or negative.

The function [tex]\( f(x) = |x + 3| \)[/tex] can be split into two cases:

1. When [tex]\( x + 3 \geq 0 \)[/tex], or equivalently [tex]\( x \geq -3 \)[/tex]:
In this case, the absolute value function becomes [tex]\( f(x) = x + 3 \)[/tex].

2. When [tex]\( x + 3 < 0 \)[/tex], or equivalently [tex]\( x < -3 \)[/tex]:
In this case, the absolute value function becomes [tex]\( f(x) = -(x + 3) \)[/tex], which simplifies to [tex]\( f(x) = -x - 3 \)[/tex].

Putting these cases together, we can rewrite [tex]\( f(x) \)[/tex] as a piecewise function:

[tex]\[ f(x) = \begin{cases} x + 3 & \text{if } x \geq -3 \\ -x - 3 & \text{if } x < -3 \end{cases} \][/tex]

Thus, the correct arrangement of the tiles and final function is:

[tex]\[ f(x) = \begin{cases} x + 3, & \text{if } x \geq -3 \\ -x - 3, & \text{if } x < -3 \end{cases} \][/tex]

This matches the solution given.