The table below contains four statements. For which function below are all four statements true? The domain of the function is x ≤ 3. The range of the function is y ≥ 0. The x-intercept of the function is 3. When x decreases, the function increases.

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.

the answer should be

d (x) = square root of 3-x.