A. If the domain is x<=3, a typical case is the square-root function, such as
f(x)=sqrt(3-x). When x=3, f(x)=0, when x=0, x=sqrt(3)...
B. testing f(x) for range of y. Since sqrt(x) function has a range of y>=0 for x real, so f(x)>=0.
C. at x=3, f(x)=f(3)=0, so satisfied.
D. f'(x)=-1/sqrt(3-x)<0 for all x, therefore when x decreases, function increases.
Therefore f(x)=sqrt(3-x) satisfies all four given conditions.
Note that this is not the only possible choice.
All functions of the form f(x)=k*sqrt(3-x) for all k>0 will satisfy the given conditions as well.
