Show that $f''(x) = 0$ for some $x$ 
Let $f$ be a twice differentiable function with the following properties: $f(x) > 0$ for $x \ge 0$. and $f$ is decreasing, and $f'(0) = 0$. Prove that $f''(x) = 0$ for some $x > 0$. 

The textbook gives the hint:

Hint: Choose $x_0 > 0$ with $f'(x_0) < 0$. We cannot have $f'(y) \le f'(x_0)$ for all $y > x_0$. 

So the alternative is: $f'(y) > f'(x_0)$
Let $x_1 > x_0$ which gives:
$$f'(x_1) > f'(x_0)$$ 
$$f''(\zeta) = \frac{f'(x_1) - f'(x_0)}{x_1 - x_0} > 0$$ 
For some $\zeta \in (x_0, x_1)$.
Since $f(0) > f(x_1)$ [because $f$ is dec.] 
$$f'(\eta) = \frac{f(0) - f(x_1)}{-x_1} < 0$$
$f'(\eta) > f'(x_0)$
But what can I do next?
 A: Since $f''$ is a derivative it satisfies the intermediate value property. If $f''$ takes both positive and negative values then $f''(x) = 0$ for some $x$. 
Suppose that $f'' < 0$  on $(0,\infty)$. Then $f'$ is strictly decreasing there. Apply the mean-value theorem: if $x > 1$ there exists $c \in (1,x)$ satisfying
$$f(x) - f(1) = f'(c)(x-1) < f'(1)(x-1).$$
Thus $f(x) < f(1) + f'(1)(x-1)$ for all $x > 1$. This forces $f(x) \to -\infty$ as $x \to \infty$, contrary to the fact that $f(x) > 0$.
Suppose that $f'' > 0$ on $(0,\infty)$. Then $f'$ is strictly increasing there, so $f'(x) > f'(0) = 0$ for all $x > 0$, contrary to the fact that $f$ is decreasing.
Thus $f''$ takes both positive and negative values, so it must vanish somewhere.
A: Taylor's expansion shows that $f(x) = f(0)+ {1 \over 2} f''(\xi) x^2$ for some
$\xi \in (0,x)$. Since $f$ is non increasing, we must have $f''(0) \le 0$.
If $f''(0) = 0$ we are finished, so suppose $f''(0)<0$.
Then there is some
$x_1>0$ such that $f'(x_1) <0$ and $f''(x_1) <0$ (so $f'$ is strictly decreasing near $x_1$). Then there must be some $x_2>x_1$ such that
$f'(x_2) = f'(x_1)$. If not, then $f'(x) < f(x_1)$ for all $x > x_1$, and
$f(x) \le f(x_1) + f'(x_1)(x-x_1)$, which contradicts $f$ being bounded below.
Then Rolle's theorem shows that there is some $\xi \in (x_1,x_2)$ such that
$f''(\xi) = 0$.
A: Just to test if mere intuition is acceptable in the case of such harmless functions I risk the following answer:
This is a continuous function that tends to zero or a positive number if x tends to the infinity. Furthermore the tangent line to this function is horizontal at zero. Because the function tends to zero or a positive number its derivative has to tend to zero. A continuous twice derivable function whose derivative is zero at zero, is decreasing and then it is flatting out has to have an inflection point somewhere after zero. This is the point. (These are the points.)
A: Since $f'$ is continuous, $f'$ restricted to $[0,x_1]$ has a minimum on this interval (by the Extreme Value Theorem). Since $f'(x_0)<0=f'(0)$ and $f'(x_0)<f'(x_1)$, this minimum does on occur on the boundaries of the interval. Therefore, $f'$ has a local minimum at some point $x^*\in(0,x_1)$, and it follows that $f''(x^*)=0$.
A: I assume that the function $f(x)$ is always positive and (strictly) decreasing for all $x>0$. 
What happen for $x$ going to infinity? Clearly, the function decreases but it remains positive. Then:
$$\lim_{x \to +\infty} f(x) = L, ~f(0) > L > 0.$$
Since $f(x)$ is differentiable, then
$$\lim_{x \to +\infty} f'(x) = 0.$$
The function $f'(x)$ is continuous and has the following properties:
$$f'(0) = 0 \\ f'(x) < 0 ~ \forall x > 0 \\ \lim_{x \to +\infty} f'(x) = 0.$$
This means that $f'(x)$ decreases just after $x=0$ but at a certain point must increase toward $0$. For the differentiability of $f'(x)$ ($f(x)$ is twice differentiable), then $f'(x)$ must have a relative minimum in $x > 0$. In other words, there must exists an $x^* > 0$ such that $$f''(x^*) = 0.$$ 
Example
Take 
$$f(x) = \frac{2+x^2}{1+x^2}, \to_{x \to +\infty} 1 $$
$$f'(x) = \frac{-2x}{(1+x^2)^2} , \to_{x \to +\infty} 0, f'(0) = 0$$
$$f''(x) = \frac{6x^2-2}{x^3+3x^4+3x^2+1}, f''\left(\frac{\sqrt3}{3}\right)=0$$
