Answer:
[tex]f(x)=3+|2x-5|=\left\{ \begin{array}{ll} 2x-2& \quad x \geq 5/2 \\ -2x+8 & \quad x < 5/2 \end{array} \right.[/tex]
Step-by-step explanation:
We are given the function:
[tex]f(x)=3+|2x-5|[/tex]
Remember that by the definition of absolute value:
[tex]\displaystyle |x|= \left\{ \begin{array}{ll} x & \quad x \geq 0 \\ - x & \quad x < 0 \end{array} \right.[/tex]
Our absolute value is:
[tex]|2x-5|[/tex]
First, we will find when it becomes 0. Set the equation equal to 0:
[tex]2x-5=0[/tex]
Solve for x:
[tex]x=5/2[/tex]
So, we can see that for all values greater than x = 5/2, 2x - 5 is positive.
For all values less than x = 5/2, 2x - 5 is negative.
Therefore (the positive case go above, and the negative case go below):
[tex]|2x-5|= \left\{ \begin{array}{ll} 2x-5 & \quad x \geq 5/2 \\ -(2x-5) & \quad x < 5/2 \end{array} \right.[/tex]
Finally, we can add a three:
[tex]f(x)=3+|2x-5|=\left\{ \begin{array}{ll} 3+(2x-5) & \quad x \geq 5/2 \\ 3+(-(2x-5)) & \quad x < 5/2 \end{array} \right.[/tex]
Simplify if desired:
[tex]f(x)=3+|2x-5|=\left\{ \begin{array}{ll} 2x-2& \quad x \geq 5/2 \\ -2x+8 & \quad x < 5/2 \end{array} \right.[/tex]
