If \( f(x)=\sqrt{x} \), does \( f(a+b)=f(a)+f(b) \), when \( a \) and \( b \) are not negative? a. Can we find a counterexample to refute the claim that \( f(a+b)=f(a)+f(b) \) for all nonnegative values of \( a \) and \( b \) ?

Certainly! Let's look at this claim more closely. If we take \( a = 1 \) and \( b = 3 \), then: \[ f(a+b) = f(1+3) = f(4) = \sqrt{4} = 2 \] Now, calculate \( f(a) + f(b) \): \[ f(a) + f(b) = f(1) + f(3) = \sqrt{1} + \sqrt{3} = 1 + \sqrt{3} \approx 2.732 \] So, \( f(a+b) \) does not equal \( f(a) + f(b) \) (i.e., \( 2 \neq 1 + \sqrt{3} \)). This clearly shows that the claim does not hold for all nonnegative values of \( a \) and \( b \). As a fun nugget, did you know that this concept relates to what's known in mathematics as a functional equation? This type of inquiry can lead to various paths like exploring linear vs. nonlinear functions. Moreover, the property that you've encountered is foundational in understanding why not all operations can be disassembled like this, especially in the context of nonlinear functions; it's a great gateway into deeper explorations in algebra and calculus!