Let the function f: R R be defined by f(x)= xe x (x2023+2024x+2025 ) (x2-x+3 )+ 2e x (x2023+2024x+2025 ) (x2-x+3 ). Then the number of solutions of f(x)=0 in R is .
Answer
1
Solution
f(x)= ( x+2) (x2023+2024x+2025 )e x (x2-x+3 ).
x+2>0, e x>0, x2-x+3>0 for all real x.
So f(x)=0 h(x)=x2023+2024x+2025=0; h'(x)=2023x2022+2024>0, so h is strictly increasing and has exactly one real root (x=-1).
Answer: 1