If $f(0) = 1$ and $f'(0) = -1$ and $f(x) > 0$ always, what can be said about $f''(x)$ Suppose we have a function $f: [0,\infty) \to R$ so that $f(0) = 1$ and $f'(0) = -1$ and $f(x) > 0$ always, what can be said about $f''(x)$ ?
I think we can't say much about $f''$ because there is no restriction but my answer book says $f''(x) < 0$ for all $x$. I think this is not possible at all, as it would mean $f(x) = 0$ for some $x \in (0,1)$.  Help me.
Also I tried considering $f(x) = \exp(-x)$
 A: You can pretty much fix any value you want for $f''(0)$.
For instance $g(x)=1-x+\frac a2x^2$ verifies $\begin{cases}g(0)=1\\g'(0)=-1\\g''(0)=a\end{cases}\quad$ for any $a\in\mathbb R$.
By continuity the condition $g(x)>0$ is true in a neighbourhood of $0$, but since $g$ is polynomial it grows to infinity at infinity, we have to make it quickly decreasing in order to build a suitable $f$.
This is possible by multiplying $g$ by $\exp(-\alpha x^4)$ (I choose $x^4$ in order to have no problem with negative numbers, and also such that the second derivative in not impacted by the value of $\alpha$).
Finding a good $\alpha(a)$ so that $f(x)\neq 0$ can be done empirically, but anyway the exponential will always win (just make the coefficient bigger).
For instance $f(x)=(1-x+\frac a2x^2)\exp\left(-\max(1000,a^4)x^4\right)$ is working well.
And $f^{(i)}(0)=g^{(i)}(0)$ for $i=0,1,2$ along with $f(x)>0$ for all $x\in\mathbb R$.
And this construction can be extended to fix any $n$-th derivative in $0$, provided you take care of the growth of the function far away of $0$ by multiplying by a suitable quickly decreasing exponential.
A: You're right. They must have omitted some important hypothesis or who knows what. Note also that the answer is not $f''(x)>0$ either. While that could be the case (as in the example you gave), you could have a change of convexity/concavity.
For instance, consider $f(x)=\frac1{1+(1+2x)^2}$. 
One could add some trigonometric functions to the mix and obtain a function which changes the sign of $f''$ indefinitely. You can say, nevertheless, that is not true that $f''(x)<0$ always. ;)
A: I think that, assuming the following, it can be inferred that $f''(x) < 0$ for most, if not all $x \in [0,1)$:


*

*The function is continuous in $[0,1)$.

*$f'(x)<0$ for $[0,1)$.


If so, it can be inferred that also $f''(x)<0$. This way it will remain with a downward slope, as well as a concave, allowing the downward slope to continue indefinitely.
I think continuousness is a fair assumption if you're usually studying that type of functions.
