Recent content by Karol

Thank you all

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...

Yes, with the Intermediate Value Theorem: So i even don't need to prove y decreases (by proving y'<0)

$$y'=6x^2 - 6x -12$$ $$\left\{ \begin{array}{l} x_1+x_2=-\frac{b}{a}=1 \\ x_1\cdot x_2=\frac{c}{a}=(-2) \end{array}\right.$$ $$x_1=(-1),~x_2=2$$ 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

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''

Y'=0 at a=(-1) How do i prove y'≠0 on the rest of the open domain <a-b>

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...

$$y(max)=45^0+2\pi k,~y(min)=225^0+2\pi k$$

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

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

$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2,~~R''=C-2D,~~R''=0:~D=\frac{C}{2}$$ 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

Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2$$ $$R'=0~\rightarrow~D=C$$

Thank you Ray, you are correct, this is only a minimum problem. I just automatically wrote the heading. Thanks

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