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
Homework Statement
Homework Equations
Minimum/Maximum occurs when the first derivative=0
The Attempt at a Solution
$$Q=\sqrt{\frac{2(K+pQ)}{h}}~\rightarrow~Q=\frac{2}{h}(KM+pM)$$
##Q'=0~## gives no sense result
a are cents absorbed by the firm. P is the price presented to the customers, is what they are asked to pay
$$\left\{ \begin{array}{l} P-(10-a)=200-0.01x \\ y=(50+a)x+20,000 \\ N=xP-y \end{array}\right.$$
$$N'=0:~0.01x=80-a$$
I don't seem to get rid of a so i will use the result for the original...
$$\left\{ \begin{array}{l} y=60x+20,000 \\ P=200-0.01x \\ N=(P-a)x-y \end{array} \right.$$
$$N=(140-a)x-0.01x^2-20,000$$
$$N'=0~\rightarrow~140-a=0.02x$$
I have to express x in terms of a, x=x(a)
##y = (50 + 10 - a)x + 20,000~## says (10-a) is absorbed by the firm. every item is more expensive to "produce" by (10-a). i shouldn't write ##~N=(P-a)x-y~## since the loss to the firm has already been taken into account in y.
So a is passed to the customer. every item is more expensive by a. if...
The answer to the original question, no' 7, is x=7500. If "all other features are unchanged" then x=7500.
$$\left\{ \begin{array}{l} P=200-0.01x-a \\ y=(60-a)x+20,000 \\ x=7500 \end{array} \right.$$
$$N'=0~\rightarrow~x=100(70-a)~\rightarrow~7500=100(70-a)~\rightarrow~a=-5$$
$$\left\{ \begin{array}{l} P=200-0.01x-a \\ y=(60-a)x+20,000 \end{array} \right.$$
$$N=xP-y=...=2(70-a)x-0.01x^2-20,000$$
$$N'=2(70-a)-0.02x,~~N'=0~\rightarrow~x=100(70-a)$$
It still is with 2 variables.
And i see that ##~P=200-0.01x~## represents the ability of the firm to charge the price P if...
Why it's not correct that: ##~y=(50+a)x+20,000~##? it's more expensive to make an item.
I don't think you can modify ##~P=200-0.01x~## because it depends only on the customers that are ready to buy x items if the price is P.
The formula ##~P=200-0.01x~## doesn't depend at all on the firm, only...
x is the same, items sold per week, and A is the new coefficient instead of 0.01.
A<0.01 since P must be higher now.
But maybe P=B-0.01x, i don't know.
Or may be even P=B-Ax
Homework Statement
Homework Equations
Minimum/Maximum occurs when the first derivative=0
The Attempt at a Solution
$$y=20,000+60x,~~P=200-Ax$$
$$N=xP-Y=200x-Ax^2-20,000-60x,~~N'=140-2Ax$$
Two variables