Relative Extrema & Points of Inflection

For 1 and 2, take the derivative and find the solution to df(x) = 0 where df(x) is the derivative of f wrt x.

For 3, I'm probably too rusty to help you. I believe a point of inflection has the slope (df(x)) increasing in one direction and decreasing in the other. So, if that's true, you would find your relative extrema and plug some values in either direction to find out if you have a min, max, or point of inflection. I think there's also a shortcut. Taking the second derivative and finding the solutions to d(df(x)) = 0 gets you something, but I can't remember what. Might be time to break out the calculus book.

Take the derivative of each polynomial function. Set the derivative to zero and solve for x (you should be able to factor these, but if not use the quadratic formula)

The first one is a parabola and has an absolute extrema. The other two are cubics and have local extremes or inflection points (freeducks definition is right).

f(x)=a(0)*x^n+a(1)*x^(n-1)+....+a(n-1)*x+a(n)

where a(0), a(1),.....a(n-1), a(n) are constants

df(x)/dx= a(0)*n*x^n-1+a(1)*(n-1)*x^(n-2)+....+a(n-1)*x

you can check for maxima or minima by taking the second derivative of the polynomial (d^2f(x)/dx^2). If the second derivative is negative at the point where the first derivative is zero it is a maxima, positive a minima, zero an inflection point.

Rap

Rap

Ah, thanks, that's the part I was trying to remember.

Are they the correct answers?
1.
f(x)=x2-6x-3
f'(x)=2x-6
x-3=0
x=3
32-(6)(3)-3=-18
Min(3,-12)
2.
F(x)=12+9x-3x2-x3
f'(x)=-3x2-6x+9
-3x2-6x+9=0
- x2-2x+3=0
-(x-1)(x+3)=0
(-x+1)=0
(x+3)=0
x=1
x=-3
Where x=1: 12+9-3-1=17
Where x=-3: 12+ (9*-3)-(3*9)-27=-69
Max(1, 17)
Min(-3, -69)
3.
f(x)=x3-6x2+12x-6
f'(x)=3 x2-12x+12
f"(x)=6x-12
6x-12=0
x-2=0
x=2
23-(6*22)+(12*2)-6=
8-24+24-6=2
Inflection points: (2, 2)

1&2 look good, but the solution to 3 looks right for a wrong reason.

Look at the first derivative
f'(x)=3x^2-12x+12
set this equal to zero (no slope--horizontal)
3x^2-12x+12=0
or
x^2-4x+4=0
(x-2)(x-2)=0
so @ x=2 f(x) has no slope

Now take the second derivative
f"(x)=6x-12
and substitute in x=2
f"(2)=6*2-12=0
this means at that as x=2, where there is no slope, the function f(x) is at an inflection point. Not that the point of no slope can be found from the second derivitive.

Rap

Thank You.
I really appreciate your help. It is my last class before I get my degree and I'd hate to blow it. Would you be willing to check a few more?

5) (3.3) find the limit , if it exists. Answer = 2

6) (3.3) find the limit , if it exists. Answer = -2

7) (3.4) find the absolute maximum and the absolute minimum values of the function f(x)=x2-4x+5, if they exist, over the interval [1, 3]
f(x)= x2-4x+5
f'(x)=2x-4
2x-4=0
x-2=0
x=2
4-8+5=1
(2, 1)
[1, 3]
X=2:4-8+5=1
1-4+5=2
1-4+5=2
X=1: 9-12+5=2
Min=1 @ x=2
Max=2 @ x=1

8) (3.4) find the absolute maximum and the absolute minimum values of the function f(x)=8x-x2, if they exist, over the real line (-∞, ∞).
f(x)=8x- x2
f'(x)=8-2X
8-2X=0
4-X=0
X=-4
8*-4-16=-48
(-4, -48)
(-∞, ∞)
Max= 4 @ x=-4

9) (3.5) Of all numbers whose difference is 6, find the two that have the minimum product. Answer = 7 & 1

10) (3.5) what is the maximum area in m2 of a rectangular ranch can 34 meters of fencing enclose? Answer = 72m^2

12) (3.6) a supply function for a certain product is given by S(p)=0.08p3+2p2+10p+11. When the price changes from $18.00 per unit to $18.20 per unit, how many more units will a seller supply approximately?

f (18) = 1305.56
f (18.2) = 1337.76544
Answer = 32

17) (4.4) How old is an ivory tusk that has lost 40% of its carbon-14? (use k=0.0001205, t in years) . Answer = ~4200 years

18) (4.6) for the demand function x=D(p)=500-p, find the elasticity.
Answer = -1/500-p

19) (4.6) based on the elasticity at p=$38.00 in above problem, is the demand elastic or inelastic?
Answer = inelastic

20) (4.6) for the same problem above, find the value of price p for which the total revenue is a maximum.
Answer = $250

5 & 6) need the functions

7) Great so far, but I'm going to try something a little different. Thing of the shape of f(x)=x^2-4x+5,. This is a parabola right? Now think about where the legs are pointed if x goes negative the x^2 term goes positive so f(-x) goes up, And if x is positive and greater than 4 it too goes up. So the max (& possibly the min) will be at one end of the interval.

So I'd first look at the endpoints,
f(x)= x^2-4x+5 @ x=1 & 3
f(1)=1-4+5=2
f(3)=9-12+5=2


Sumpting tells me that the minimum is somewhere between 1 and 3
So now I would look at the derivative of f(x) (& set it equal to 0)
f'(x)=0
Which you did correctly
@ x=2 f'(2)=0
plug x=2 into f(x) &
f(2)=4-8+5=1

now remember this shape is a upward parabola with a minimum @ X=2 & f(2)=1, and equal maxes @ x=1, f(1)=2 and x=3, f(3)=2

I see the line of symmetry of the parabola being at the minimum, with 1 and 3 being equidistant from 2, and legs pointing upward, the ends are equal maximums.

So the answer is
Max @ x=1,3 and minimum @ x=2

8) (3.4) find the absolute maximum and the absolute minimum values of the function f(x)=8x-x2, if they exist, over the real line (-∞, ∞).

Again think of the shape of this equation f(x)=-x^2-8x, think of it as f(x)=-x^2-8x+0
Two things you should know immediately, it's a parabola and f(0)=0. Now look at direction of the ends. As you send x toward ∞ f(x) goes to negative ∞. So this sucker has legs pointed down' so the minimums are easy & there are two of them x->∞ & x->-∞

Now look for the max & your right

Now go back to the shape of the parabola and think about what it looks like.

9) (3.5) of all numbers whose difference is 6, find the two that have the minimum product.
I will assume that the numbers in question are positive counting numbers, if not the answer is 0 and 6, since 6-0=6 and 6*0=0

So I gotta think about counting numbers and the product of counting numbers. Now remember that anything multiplied by 1 is itself. So 1 is one of the counting numbers. And A-1=6, so A=7

Interestingly, when I did this as a continuous function I kept coming up with 3 and -3
As a max or min.

Answer = 7 & 1

10) (3.5) what is the maximum area in m2 of a rectangular ranch can 34 meters of fencing enclose? Your answer is wrong
Think of a rectangle
The perimeter is twice the sum of the width and length
The area is the product

P=2(l+w)
A=lw

find where P is a minimum and A is max
dP=0=2(dl+dw) so dl/dw=-1
dA=0=wdl+ldw so dl/dw=-l/w
if -l/w=-1 the l=w
if l=w P=4l=34 so l=34/4=17/2 and A=(17/2)^2m^2

12) (3.6) a supply function for a certain product is given by S(p)=0.08p3+2p2+10p+11. When the price changes from $18.00 per unit to $18.20 per unit, how many more units will a seller supply approximately?

No reason to carry the fraction and I'd round the function values

f (18) = 1305.56
f (18.2) = 1337.76544
Answer = 32

17) (4.4) How old is an ivory tusk that has lost 40% of its carbon-14? (use k=0.0001205, t in years) .

C-12 has a half life of 1600 years or so iff'n I remember. So the decay constant (λ) is ln(2)/5500 yr^(-1) ------I like units in physics
This is about λ=1.205E-4 yr---I also like scientific notation
& 0.4=e^(- λt)
yer fergetting to take the natural logarithm
ln (0.4)= - λt
& I don't get 4200 years

Answer = ~4200 years

I don't understand 18, 19, and 20)

Rap

Answers
Hey Rap.
These are the answers.
1. Min(3,-12)
2. Max(1, 17), Min(-3, -15)
3. x=2, where there is no slope, the function f(x) is at an inflection point
4. C
5. -3
6. Does not exist
7. Max @ x=-1 Min @ x=2
8. Max @ x=4 No minimum
9. 3 & -3
10. 72m^2
11. B
12. 32
13. A
14. A
15. C
16. B
17. ~4200 years
18. C
19. inelastic
20. $250

mea culpa--I now get 4200 years

Rap