Answer:
m=1 and n=3
Step-by-step explanation:
One way of doing this is solving the differential equation of f(x).
We have that [tex]f'(x)=f(x)[/tex] so [tex]f'(x)/f(x)=1[/tex], that is, [tex](\ln(f(x))'=1[/tex]. Integrating in both sides respect to x, [tex]\ln(f(x))=x+C[/tex] for some constant C. Therefore [tex]f(x)=e^{x+C}=ke^x[/tex] for all x, and for some constant [tex]k=e^C[/tex]. In particular, taking x=3, [tex]f(3)=ke^3=e^2[/tex]. From this, [tex]k=e^{-1})[/tex], then [tex]f(x)=e^{x-1}[/tex]. Choosing x=4, we have that [tex]f(4)=e^3[/tex] so the integers are m=1 and n=3.
