\( e ^ { f ( x ) } - 1 = f ^ { 2 } ( x ) \)

You’ve provided the equation: \[ e^{f(x)} - 1 = [f(x)]^2 \] To solve for \( f(x) \), let's analyze the equation step by step. ### Step 1: Understand the Equation \[ e^{f(x)} - 1 = [f(x)]^2 \] This is a transcendental equation because it involves both an exponential function and a polynomial. Solving such equations analytically can be challenging, but in this case, we can find real solutions by analyzing the behavior of the functions involved. ### Step 2: Let’s Set \( y = f(x) \) The equation becomes: \[ e^{y} - 1 = y^2 \] We need to find the values of \( y \) that satisfy this equation. ### Step 3: Analyze for Real Solutions Consider the function: \[ h(y) = e^{y} - 1 - y^2 \] We want to find the roots of \( h(y) = 0 \). #### Testing \( y = 0 \): \[ h(0) = e^{0} - 1 - 0^2 = 1 - 1 - 0 = 0 \] So, \( y = 0 \) is a solution. #### Checking for Other Real Solutions: - **For \( y > 0 \):** The exponential function grows faster than the quadratic function. For example: - At \( y = 1 \): \[ h(1) = e^1 - 1 - 1^2 \approx 2.718 - 1 - 1 = 0.718 > 0 \] - At \( y = 2 \): \[ h(2) = e^2 - 1 - 2^2 \approx 7.389 - 1 - 4 = 2.389 > 0 \] Since \( h(y) \) is increasing for \( y > 0 \) and \( h(0) = 0 \), there are no other positive real solutions. - **For \( y < 0 \):** The exponential function approaches zero, while the quadratic term becomes positive. - At \( y = -1 \): \[ h(-1) = e^{-1} - 1 - (-1)^2 \approx 0.368 - 1 - 1 = -1.632 < 0 \] - As \( y \) approaches negative infinity, \( e^{y} \) approaches 0, and \( y^2 \) becomes very large. Therefore, \( h(y) \) tends to negative infinity. The function \( h(y) \) increases as \( y \) approaches 0 from the negative side but does not cross zero again because: \[ \frac{dh}{dy} = e^{y} - 2y \] At \( y = 0 \), \( \frac{dh}{dy} = 1 > 0 \), indicating that \( h(y) \) is increasing at \( y = 0 \). Given that \( h(y) \) approaches negative infinity as \( y \) approaches negative infinity and is negative near \( y = -1 \), there are no other real roots besides \( y = 0 \). ### Step 4: Conclusion The only real solution to the equation \( e^{f(x)} - 1 = [f(x)]^2 \) is: \[ f(x) = 0 \] ### Verification Plugging \( f(x) = 0 \) back into the original equation: \[ e^{0} - 1 = 0^2 \implies 1 - 1 = 0 \implies 0 = 0 \] The equation holds true. ### Note on Complex Solutions While we've found the only real solution, there may be complex solutions to the equation. Solving for complex \( y \) would require more advanced techniques, including using the Lambert W function. However, unless you're specifically interested in complex solutions, \( f(x) = 0 \) is the primary solution.