I tried solving a calculus problem and I got the right result, but I don't understand the solution provided at the end of the exercise. Even though I got the same answer, I would like to understand what's happening in the given solution aswell.
Consider the function: $$\ f(x) = \begin{cases} x^2+ax+b & x\leq 0 \\ x-1 & x>0 \\ \end{cases} \ $$ Find the antiderivatives of the function $f$ if they exist.
The solution provided goes something like this:
For $f$ to have antiderivatives the function $f$ must have the Darboux property. (...Some calculations...), therefore $f$ has the Darboux property if and only if $b = -1$ (I understood that now the function is continuous, therefore it has an anitederivative). Using the consequences of Lagrange's theorem on the intervals $(-\infty, 0)$ and $(0, \infty)$ any antiderivative $F : \mathbb{R} \rightarrow \mathbb{R}$ of $f$ has the form:
$$ F(x) = \ \begin{cases} \dfrac{x^3}{3} + a \dfrac{x^2}{2} - x + c_1 & x < 0 \\ \ c_2 & x=0 \\ \dfrac{x^2}{2} - x + c_3 & x>0 \end{cases} \ $$
$F$ being differentiable, it is also continous, so $F(0) = c_2 = c_1 = c_3 $.
Therefore the antiderivatives of $f$ have the form:
$$ F(x) = c + \ \begin{cases} \dfrac{x^3}{3} + a \dfrac{x^2}{2} - x & x\leq 0 \\ \dfrac{x^2}{2} - x & x>0 \end{cases} \ $$
Again, I got the same result, but I don't understand a lot of the work done above.
The first thing I didn't understand is the part where they say that $f$ has an antiderivative iff it has the Darboux property. I searched a bit online and I found that a function accepts antiderivatives only if it has the Darboux property. So I guess I have to accept that as a fact.
The second (and more important thing) that I didn't understand was the part where they said that they used the consequences of Lagrange's Theorem on the intervals $(-\infty, 0)$ and $(0, \infty)$ to find that first form of the antiderivative. What theorem are they refering to? How did they use it on those intervals? Why is there a separate case for $x = 0$ with an aditional constant, $c_2$. I only used $2$ constants, why were there needed $3$? Long story short, I just don't understand at all how they arrived at that first form of the antiderivative and how they used these "consequences of Langrange's theorem". I understood the second form of the antiderivative, that's what I also got, but the first form put me in the dark.
I know these are all just details, but I really want to understand what was used here, why was it used and how was it used.
from Hot Weekly Questions - Mathematics Stack Exchange
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