1. (extra credit 10 pts.) Consider the function of two variables given by \( f(x, y)=\frac{1}{2}(x+y+|x-y|) \) Experiment with \( f \) on scratch paper for various real-number inputs in place \( x \) and \( y \). Show two of the calculations below and briefly explain what this function does.

1. Calculation for \( x = 5 \) and \( y = 3 \): - Compute the absolute difference: \( |5 - 3| = 2 \) - Substitute into the function: \( f(5, 3) = \frac{1}{2}\left(5 + 3 + 2\right) = \frac{1}{2}(10) = 5 \) - Since \( 5 \) is the larger number, the function returns \( 5 \). 2. Calculation for \( x = -4 \) and \( y = 2 \): - Compute the absolute difference: \( |-4 - 2| = |-6| = 6 \) - Substitute into the function: \( f(-4, 2) = \frac{1}{2}\left(-4 + 2 + 6\right) = \frac{1}{2}(4) = 2 \) - Here, the larger number is \( 2 \), and the function returns \( 2 \). **Explanation:** In both cases, after simplifying the expression, the function \( f(x, y) = \frac{1}{2}(x+y+|x-y|) \) always returns the greater of \( x \) and \( y \). This is because when \( x \ge y \): \[ |x-y| = x-y \quad \Longrightarrow \quad f(x,y)=\frac{1}{2}(x+y+x-y)=x, \] and when \( y > x \): \[ |x-y| = y-x \quad \Longrightarrow \quad f(x,y)=\frac{1}{2}(x+y+y-x)=y. \] Thus, the function effectively computes the maximum of the two values.