Question

50 POINTS FIRST GOOD ANSWER GETS BRAINLIEST

Answer 1

Answer:

$f(x)=3+|2x-5|=\left\{ \begin{array}{ll} 2x-2& \quad x \geq 5/2 \\ -2x+8 & \quad x < 5/2 \end{array} \right.$

Step-by-step explanation:

We are given the function:

$f(x)=3+|2x-5|$

Remember that by the definition of absolute value:

$\displaystyle |x|= \left\{ \begin{array}{ll} x & \quad x \geq 0 \\ - x & \quad x < 0 \end{array} \right.$

Our absolute value is:

$|2x-5|$

First, we will find when it becomes 0. Set the equation equal to 0:

$2x-5=0$

Solve for x:

$x=5/2$

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):

$|2x-5|= \left\{ \begin{array}{ll} 2x-5 & \quad x \geq 5/2 \\ -(2x-5) & \quad x < 5/2 \end{array} \right.$

Finally, we can add a three:

$f(x)=3+|2x-5|=\left\{ \begin{array}{ll} 3+(2x-5) & \quad x \geq 5/2 \\ 3+(-(2x-5)) & \quad x < 5/2 \end{array} \right.$

Simplify if desired:

$f(x)=3+|2x-5|=\left\{ \begin{array}{ll} 2x-2& \quad x \geq 5/2 \\ -2x+8 & \quad x < 5/2 \end{array} \right.$