<s>[INST] Unclear calculus problem I don't know if that's called calculus or precalculus in the Anglosphere, so please pardon me if I got the terminology incorrect.
I've been given the second derivative of $f(x): f''\left(x\right)=\frac{2x-1}{\sqrt{-x^2+x+6}} $.
I was asked to find $f'(x)$, if it is known that the slopes of all tangent lines to the graph of $f(x)$ in the range of $-2<x<3$ are not smaller than -1. 
I integrated the second derivative:
$$f'\left(x\right)=\int \:\frac{2x-1}{\sqrt{-x^2+x+6}}dx\:=\:-2\sqrt{-x^2+x+6}\:+\:C$$
So the task at hand is essentially to find the value of C. The problem, however, is that I did not understand how to do that. The only thing I managed to conclude from the given information (about the slopes of the tangent lines) is that $f'\left(x\right)\ge -1$, and I don't know how to use an inequality to find C.
If someone can make it clearer, I'll be extremely glad :)
P.S. The final answer (according to the textbook) is $f'\left(x\right)=\:-2\sqrt{-x^2+x+6}\:+\:4$.
 [/INST] Just continue
So
For $-2 < x < 3$ we have $-2\sqrt{-x^2+x+6} + C \ge -1$
$2\sqrt{-x^2 + x + 6} \le 1 + C$
So $ C \ge \max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)} - 1$.
So must find $\max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)}$.  As $-x^2 +x+6$ must be non-negative we may assume $\max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)}$ where $-x^2 + x + 6$ does (assuming we get a non-negative value).
So solving for $(-x^2 + x + 6)' = -2x + 1 = 0$ whe have at $x = \frac 12$ is an extrema (and as $(-x^2 + x + 6)'' = -2 < 0$ it is a maximum).
So $C \ge 2\sqrt{-\frac 14 + \frac 12 + 6} - 1$
$= 2\sqrt{6\frac 14}-1 = 2\sqrt {\frac {25}4} - 1=5-1 = 4$.
So
$f'(x) = -2\sqrt{-x^2+x+6} + C$ where $C \ge 4$.  I don't see anything in the problem that suggest $C$ can't be larger than $4$.  However if it said there were one or more tangent lines where the slope is $-1$ but none where it is less, then $C$ must equal $4$.
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