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...
$$y'=6x^2 - 6x -12$$
$$\left\{ \begin{array}{l} x_1+x_2=-\frac{b}{a}=1 \\ x_1\cdot x_2=\frac{c}{a}=(-2) \end{array}\right.$$
$$x_1=(-1),~x_2=2$$
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
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...
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
$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2,~~R''=C-2D,~~R''=0:~D=\frac{C}{2}$$
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
$$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2$$
$$R'=0~\rightarrow~D=C$$
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