Question

Is it possible for a function f(x) defined for all x to have the following three properties: f(x) > 0, f'(x) < 0, and f''(x) < 0 ?

I know that the answer is "No" but I can't seem to put what I understand into words and explain it.

Answer 1

See below.

Explanation:

`f(x)>0` implies the function itself is positive.

`f'(x)<0` implies the first derivative is negative.

`f''(x)<0` implies that the derivative is also negative.

Let's consider whether any such functions exist.

Well,

`f(x)=cos^2x>0`

`f'(x)=-2sinxcosx=-sin2x<0`

`f''(x)=-2cos2x<0`

It would appear that `f'(x), f''(x)<0` but since sine and cosine oscillate, the two derivatives are positive for some values of `x.`

Now, consider

`f(x)=e^-x>0`
`f'(x)=-e^-x<0`
`f''(x)=e^-x>0`

This meets the conditions `f(x)>0, f'(x)<0` but not `f''(x)<0.`

Let's consider why.

`f(x)>0` implies our function is never negative. `f'(x)<0` implies the function is always decreasing. Combined with the first statement, this means the function is always positive and decreasing, as with `f(x)=e^-x.`

Now, `f''(x)<0` implies the function is always concave down. Combined with the first two, it means the function is always positive, always decreasing, and concave down.

That's just not possible. A function that is always decreasing and concave down looks something like this:

graph{-e^x+20 [-10, 10, -5, 5]}

As in, it rapidly approaches `-oo`, meaning it eventually becomes negative. So, `f(x)>0, f''(x)<0` contradict one another.