find a function such that f(x) is decreasing and concave down everywhere on R(all real numbers) with f(0) = 40.

Answer 1

If f(x) is decreasing and concave down on R, that means that f'(x) < 0 and f''(x) < 0 for all x ∈ R. After a bit of thought, the first function that comes to mind is the exponential function, e^x, which, although it is constantly increasing and concave up, can be modified to solve the problem.

After multiplying by -1 to fit the conditions of the first and second derivatives, we can then add a constant to satisfy the initial condition of f(0) = 40:

f(x) = -e^x + c

f(0) = 40 = -e⁰ + c

c = 41.

So, f(x) = -e^x + 41 meets the requirements of the problem, although any base b > 1 can be used and still work.

Answer 2

40-x² should work.