1) f(x) = Vx - 2 2) f(x) = |x - 2|
It looks like you've provided two mathematical functions. Here’s a brief overview and analysis of each function:
Function 1: \( f(x) = \sqrt{x} - 2 \)
Domain: Since we are dealing with a square root function, the expression under the square root must be non-negative. Hence, \( x \geq 0 \). Therefore, the domain is \( [0, \infty) \).
Range: The square root function \( \sqrt{x} \) has a range of \( [0, \infty) \). When we subtract 2, the minimum value occurs when \( x = 0 \), giving \( f(0) = \sqrt{0} - 2 = -2 \). As \( x \to \infty \), \( f(x) \to \infty \). Thus, the range is \( [-2, \infty) \).
Key Points:
- \( f(0) = -2 \)
- \( f(1) = \sqrt{1} - 2 = -1 \)
- \( f(4) = \sqrt{4} - 2 = 0 \)
Graph: The graph starts at (-2) when \( x = 0 \) and increases continuously, approaching positive infinity.
Function 2: \( f(x) = |x - 2| \)
Domain: The absolute value function is defined for all real numbers, so the domain is \( (-\infty, \infty) \).
Range: The output of the absolute value function is always non-negative. Therefore, the range is \( [0, \infty) \).
Key Points:
- \( f(2) = |2 - 2| = 0 \)
- For \( x < 2 \) (e.g., \( x = 0 \)), \( f(0) = |0 - 2| = 2 \)
- For \( x > 2 \) (e.g., \( x = 4 \)), \( f(4) = |4 - 2| = 2 \)
Graph: The graph is a V-shape, with the vertex at the point (2, 0). As \( x \) moves away from 2, the function value increases.
Summary:
- Function 1 is a shifted square root function, defined only for non-negative \( x \) and with a minimum value of -2.
- Function 2 is a V-shaped absolute value function, defined for all real numbers and with a minimum value of 0 at ( x = 2 .
If you need further analysis or have specific questions regarding these functions, feel free to ask!
