# Recent content by Karol

1. ### Rolle's theorem, to show there's only one root

Thank you all
2. ### Rolle's theorem, to show there's only one root

Why with the Mean Value Theorem and Rolle's Theorem there is only one root (y=0) if f(a) and f(b) have opposite signs? The Mean Value Theorem says there is at least one root, what is the role of Rolle's theorem in that there is only one root? why does it make this if y'≠0? Rolle's theorem is...
3. ### Rolle's theorem, to show there's only one root

Yes, with the Intermediate Value Theorem: So i even don't need to prove y decreases (by proving y'<0)
4. ### Rolle's theorem, to show there's only one root

$$y'=6x^2 - 6x -12$$ $$\left\{ \begin{array}{l} x_1+x_2=-\frac{b}{a}=1 \\ x_1\cdot x_2=\frac{c}{a}=(-2) \end{array}\right.$$ $$x_1=(-1),~x_2=2$$ And in between y'<0 so y is decreasing, but it doesn't matter since the theorem only demands y'≠0 in the open interval
5. ### Rolle's theorem, to show there's only one root

y'=0 at x=(-2) and x=3, between them y'<0 so y is decreasing So between a=(-1) and b=0 y decreases I don't need y''
6. ### Rolle's theorem, to show there's only one root

Y'=0 at a=(-1) How do i prove y'≠0 on the rest of the open domain <a-b>
7. ### Rolle's theorem, to show there's only one root

Homework Statement Homework Equations Rolle's Theorem: If f(a)=f(b)=0 then there is at least one a<c<b such that f'(c)=0 The Attempt at a Solution $$y=2x^3-3x^2-12x-6~\rightarrow~y'=6x^2-6x-12$$ The function: y': How do i know y' isn't 0 somewhere? if it's continuously descending, so i...
8. ### Min max: y=sin x+cos x

$$y(max)=45^0+2\pi k,~y(min)=225^0+2\pi k$$
9. ### Min max: y=sin x+cos x

Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$y=\sin{x}+\cos{x}~\rightarrow~y'=\cos{x}-\sin{x}$$ $$y'=0:~\rightarrow~\cos{x}=\sin{x}~\rightarrow~x=\frac{\pi}{4}+n\cdot \pi$$ It's not correct

Thanks
11. ### Min max: optimal quantity of medicine

Yes, that's correct, i need the maximum for R', but why? At the point where R has a maximum, i think, a small change in D makes a big change in R
12. ### Min max: optimal quantity of medicine

$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2,~~R''=C-2D,~~R''=0:~D=\frac{C}{2}$$ But the greatest change in R for a small change in D is where R has a maximum, hence where R'=0, not where R''=0
13. ### Min max: optimal quantity of medicine

Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2$$ $$R'=0~\rightarrow~D=C$$
14. ### Minimization problem: Economics: quantity to order

Thank you Ray, you are correct, this is only a minimum problem. I just automatically wrote the heading. Thanks
15. ### Minimization problem: Economics: quantity to order

In the old model: $$A(Q)=\frac{KM}{Q}+cM+\frac{hQ}{2}$$ Where c is the purchase cost of one item. in the new model: $$A(Q)=\frac{K+pQ}{Q/M}+cM+\frac{hQ}{2}=\frac{KM}{Q}+(c+p)M+\frac{hQ}{2}$$ And differentiating gives the same result