Change of sign of derivative Suppose we have a function $f$ on the real line and $f > 0$ on some interval $[a,b)$ and $f(b) = 0$. Assume also that $f'<0$ on $[a,b)$ and $f'(b)=0$. Is it possible to define $f$ in such a way that $f < 0$, $f'> 0$ on some interval $(b,c]$ ?
I suppose it is not possible but I don't really know how to prove that no such function exists.
 A: The mean value theorem is your friend any time you want to pass from information about a derivative on an interval to information about the function. Here, you're assuming $f$ is differentiable on $[b, c]$. The mean value theorem guarantees there exists a $z$ in $(b, c)$ such that
$$
f'(z) = \frac{f(c) - f(b)}{c - b}.
$$
Since $f(b) = 0$ and $c > b$, the fraction on the right has the same sign as $f(c)$, so $f(c) < 0$ and $f'(z) > 0$ are incompatible.
A: So if i will see the first condition for the function then it's giving me various clue that

*

*Function is lies upper x axis in between $[a , b)$


*secondly , in between $[a ,b)$ , the derivative of function is less than 0 means function is decreasing in $[a ,b)$


*so function stats decreasing from $a$ and continuously decreasing upto point $b$ and become 0 there. Means touch the x axis at b.
Please try to make graph in mind . It's like a decreasing curve from $a$ to $b$ and become 0 at $b$ .
NOW SEE WHAT HAPPENS IN BETWEEN $b$ to $c$

*

*The graph is starting from b now and now it's going below x Axis as function Become negative. Ok it's good and no problem in it. It can happen. Right .


*But after b to c , function should it's convature And should Become increasing now as derivative of function is positive after b to c which is clearly not possible. . why let me tell you in sort
Because function us decreasing between a to b and after that it's become increasing between a to c.  And offcourse derivative at point b is 0.
So let me mathematically formulate it ;
$f^{'}(x) \lt 0  \forall [a,b)$
$f^{'}(x) \gt 0 \forall (b , c)$
So function should attain local minima at $x = b$
And recall the definition of local minima which is that
For every $x$ $\in$ $N_{\delta} (b)$ ,
$$f(b) \lt f(x)$$
Which is not possible here. You can re read the whole procedure and can try to sketch it on notebook then formulate it mathematically and you will able to proof it.
I hope you understand the concept. If have any doubt then i am waiting for your comment . Best of luck !!
A: $$f(b)=0\land \forall x>b:f'(x)>0$$ gives by integration
$$f(c)=f(b)+\int_b^c f'(x)dx>f(b)>0.$$
