If $f(x)>0$ $ \forall x$ and $f(0)=1, f'(0)=-1$
then the statement : " $f''(x)>0$ for atleast one value of $x$? " is true or false ?
I tried the following example:
Let $f(x)=\exp(-x)$, then
$$f(0)=\exp(0)=1,\ \ f'(x)=- \exp(-x) \implies f'(0)=-1.$$
So we find $f''(x)= \exp(-x) > 0$ for all $x$.
So the above statement is true when $f(x)= \exp(-x)$.
If $f(x)=2-\cos(x)-\sin(x) $
then, $$f(0)=2 - 1 - 0 =1, f'(x)= \sin (x) - \cos(x) \implies f'(0)=-1$$
Now, $f''(x)= \cos(x) + \sin(x)$ which is also $>0$ for some values of $x$
So the above statement is true when $f(x)= 2-\cos(x)-\sin(x)$.
But I don't know whether the statement is true in general or not?
Any help will be appreciated.
Let's say we don't have $f''(x)>0$ for at least one $x$; that means we have $f''(x)\leq 0$ for all $x$. What does this say about $f'(x)$? If the derivative is never positive, it means that $f'(x)$ is non-increasing. What does this in turn say about $f(x)$? If the derivative is never increasing, it means that $f(x)$ is concave.
Try drawing $f$ around $x=0$. You know $f$ is concave, $f(0)=1$ and even the slope at $x=0$ is given; $f'(0)=-1$. Do you think we can "prevent" $f$ from crossing the x-axis and violating the $f(x)>0$ requirement? Why (not)?