The problem asks us to evaluate three functions at $x=2$: a. $f(x) = \sqrt{3x-7}$ b. $f(x) = \frac{x+1}{2x-4}$ c. $f(x) = |2x+3|$

1. Problem Description

The problem asks us to evaluate three functions at $x=2$:

a. $f(x) = \sqrt{3x-7}$

b. $f(x) = \frac{x+1}{2x-4}$

c. $f(x) = |2x+3|$

2. Solution Steps

a. $f(x) = \sqrt{3x-7}$

Substitute $x=2$ into the function:

$f(2) = \sqrt{3(2)-7} = \sqrt{6-7} = \sqrt{-1}$

Since the square root of a negative number is not a real number, the function is undefined at $x=2$.

b. $f(x) = \frac{x+1}{2x-4}$

Substitute $x=2$ into the function:

$f(2) = \frac{2+1}{2(2)-4} = \frac{3}{4-4} = \frac{3}{0}$

Since division by zero is undefined, the function is undefined at $x=2$.

c. $f(x) = |2x+3|$

Substitute $x=2$ into the function:

$f(2) = |2(2)+3| = |4+3| = |7| = 7$

3. Final Answer

a. $f(2) = \sqrt{-1}$ (Undefined or not a real number)

b. $f(2) = \frac{3}{0}$ (Undefined)

c. $f(2) = 7$