Certainly!
To rewrite the absolute value function [tex]\( f(x) = -4|x+1| + 2 \)[/tex] as a piecewise function, we need to consider the definition of the absolute value function, which is given by:
[tex]\[ |x+1| =
\begin{cases}
x + 1 & \text{if } x + 1 \geq 0 \\
-(x + 1) & \text{if } x + 1 < 0
\end{cases} \][/tex]
This means we have two cases to consider based on the value of [tex]\( x \)[/tex]:
1. Case 1: [tex]\( x + 1 \geq 0 \)[/tex]
This inequality can be simplified to:
[tex]\[ x \geq -1 \][/tex]
In this case, the absolute value function [tex]\( |x+1| \)[/tex] can be replaced by [tex]\( x + 1 \)[/tex]. Therefore,
[tex]\[ f(x) = -4|x+1| + 2 = -4(x+1) + 2 \][/tex]
Simplifying the expression:
[tex]\[ f(x) = -4x - 4 + 2 \][/tex]
[tex]\[ f(x) = -4x - 2 \][/tex]
2. Case 2: [tex]\( x + 1 < 0 \)[/tex]
This inequality can be simplified to:
[tex]\[ x < -1 \][/tex]
In this case, the absolute value function [tex]\( |x+1| \)[/tex] can be replaced by [tex]\( -(x + 1) \)[/tex]. Therefore,
[tex]\[ f(x) = -4|x+1| + 2 = -4(-(x+1)) + 2 \][/tex]
Simplifying the expression:
[tex]\[ f(x) = -4(-x - 1) + 2 \][/tex]
[tex]\[ f(x) = 4x + 4 + 2 \][/tex]
[tex]\[ f(x) = 4x + 6 \][/tex]
So, the original absolute value function can be rewritten as a piecewise function:
[tex]\[ f(x) =
\begin{cases}
-4(x + 1) + 2 \quad \text{if } x \geq -1 \\
-4(-(x + 1)) + 2 \quad \text{if } x < -1
\end{cases} \][/tex]
Putting this all together, we get the piecewise function:
[tex]\[ f(x) =
\begin{cases}
-4x - 2 \quad \text{if } x \geq -1 \\
4x + 6 \quad \text{if } x < -1
\end{cases} \][/tex]
Hence, the function [tex]\( f(x) = -4|x+1| + 2 \)[/tex] expressed as a piecewise function is:
[tex]\[ f(x) =
\begin{cases}
-4(x + 1) + 2 \quad \text{if } x \geq -1 \\
-4(-(x + 1)) + 2 \quad \text{if } x < -1
\end{cases} \][/tex]
