2. Let \( f(x)=x+e^{x}-2=0 \) has a root in the interval \( (0,1) \). If the fixed point iteration method is used which one is the iteration formula? \( \begin{array}{llll}\text { a. } g(x)=2-e^{x} & \text { b. } g(x)=\ln (2-x) & \text { c. } g(x)=\frac{x}{x+e^{x}} & \text { d. } g(x)=\frac{x e^{x}}{1-x} \\ \text { e. } g(x)=1-e^{x} & \end{array} \)

To find a suitable iteration formula using the fixed point iteration method, we need to rearrange the equation \( f(x) = 0 \) so that \( x \) is expressed in terms of \( x \). Starting from the equation \( f(x) = x + e^{x} - 2 \), we can isolate \( x \). A good approach is to express \( x \) in terms of the other parts of the equation. Looking at the proposed options, let's see which one can logically stand as a rearranged form of \( f(x) = 0 \): 1. **Option a: \( g(x) = 2 - e^{x} \)** can be obtained from \( x = 2 - e^{x} \), which is suitable. Thus, the iteration formula suitable for fixed point iteration is **a.** \( g(x) = 2 - e^{x} \). 2. **Option b: \( g(x) = \ln(2 - x) \)** does not relate directly to the equation derived and might not converge. In summary, the iteration formula derived from \( f(x) = 0 \) that would allow us to find the root in the interval \( (0, 1) \) in a fixed point iteration context is **a. \( g(x) = 2 - e^{x} \)**.