Question

Find Extreme Values of x^4-4x^3-x^2+12x-2

Answer 1

I have my answers =, but im not sure if they are right

Answer 2

what did you get for your answer, i'll check

Answer 3

max:(1.14,6.14) min: (-.94,-10), (2.8,-2.6)

Answer 4

yep, good work ^_^

Answer 5

Those are the correct extreme values? Great!

Answer 6

Would you mind checking another one or two of my answers?

Answer 7

sure thing

Answer 8

Post here or new post?

Answer 9

just here is fine

Answer 10

Use Descartes' rule of signs to describe the roots of h(x)= 4x^4-5x^3+2x^2-x+5? I have four real positive roots and 3 negative real roots

Answer 11

mmm.... i'm not familiar with this rule, sorry...

Answer 12

ok..., write a function with the given characteristics.
A polynomial with rational coefficients having roots 3, 3, and 3-i

My answer is:x^3-9x^2+28x-30

Answer 13

what grade are you in?

Answer 14

12?

Answer 15

I'm in pre calc

Answer 16

Hey @ranga can you check my last two answers?

Answer 17

@washcaps ya, i'm happy I even got you first question, haha
good luck!

Answer 18

Thanks man!

Answer 19

For the second one, it is a fourth degree polynomial. Therefore it will have four roots in all. It cannot have four real positive roots AND 3 negative real roots.

Answer 20

hmm so 2 real positive and 2 negative? I think there can be 4,2,or 0 real positive and 3 or 1 negative

Answer 21

4x^4-5x^3+2x^2-x+5
The sign changes 4 times. Max positive roots is:4.
The possibilities for positive roots are: 4, 2, 0

replace x by -x and count the sign changes.

Answer 22

2 positive and 1 negative?

Answer 23

If you replace x by -x in 4x^4-5x^3+2x^2-x+5 you get:
4(-x)^4 - 5(-x)^3 + 2(-x)^2 - (-x) + 5
4x^4 + 5x^3 + 2x^2 + x + 5
No sign changes. Therefore there are NO negative roots.

So to summarize, these are the possibilities for the roots of 4x^4-5x^3+2x^2-x+5:

Positive Complex
4 0 OR
2 2 OR
0 4

Answer 24

Oh crap haha so there are just four real positive roots. So that's my answer

Answer 25

No, the maximum is 4.
It can be 2 positive roots or NO positive roots also.
I have listed the three possibilities above.

Answer 26

But there is just one answer right?

Answer 27

No you have to list all three possibilities.
That is what the Descartes rule of signs tell us. They tell what the possibilities are.

Answer 28

Really? for the question se Descartes' rule of signs to describe possible combinations of roots for the polynomial.

2x^3+5x^2-31x-15

I put: One real positive root and two negative real roots. I got 100 on the assignment

Answer 29

It has definitely only ONE positive root.
But for the negative roots the possibilities are: 2 or 0.
Since this is a cubic function, it will have a total of 3 roots.

So the correct way to answer is to list all possibilities:

Positive Negative Complex
1 2 0
1 0 2

By the way for the second problem if you really find the roots of 4x^4-5x^3+2x^2-x+5 you will see that it has NO real roots (positive or negative). All 4 roots are complex! And we have listed that as one of the possibilities.

Answer 30

And you are 100% sure? It is worth a lot of points. It says describe the roots not the possible roots

Answer 31

I don not know what your teacher wants or expects. But the Descartes rule of signs gives you the possibilities. Only when you actually find the roots you will know which possibility is true. Until then all you can say it has to be one of these possibilities.

Go to YouTube and type Descartes rule of signs and see some of the examples to see how others are doing it and then you decide how you want to answer the question.

Answer 32

If we actually solved the roots, what would be the answer

Answer 33

o positive and 4 complex roots?

Answer 34

For the second problem ALL roots are complex. No real roots at all.
For the third problem, 2 negative and one positive.

But if you are going to be finding the actual roots to answer the question then there is no reason to do Descartes rule of signs at all.

Answer 35

Oh ok. Did you check my other question, the 3rd one?

Answer 36

I though the third problem was: 2x^3+5x^2-31x-15

Answer 37

no that was already done. I asked the other question right before you came in

Answer 38

can you repeat the question here that you want me to take a look.

Answer 39

write a function with the given characteristics.
A polynomial with rational coefficients having roots 3, 3, and 3-i

My answer is:x^3-9x^2+28x-30

Answer 40

They show you two real roots and one complex.
Always remember that complex roots ALWAYS occur in pairs.
They are called conjugates.
So if 3-i is a root, then 3+i will also be a root.
So there are 4 roots here: 3, 3, 3-i and 3+i

Answer 41

No it's not asking for roots... it is telling me to make a function from the roots given. Is my function correct

Answer 42

Yes. But the point I am making is there are 4 roots. Therefore, the answer will be a 4th degree polynomial.

Since you know the four roots you can find the polynomial which is the product of:
(x-3) * (x-3) * {x-(3-i)} * {x-(3+i)}

Multiply the last two parts first to get rid of the i. Then multiply the rest and arrange the polynomial from the highest degree to the lowest.

Answer 43

But my degree is only 3

Answer 44

why do you say the degree must be 3?

Answer 45

Not must be three. But my function I asked you to check has a degree of 3. You said mine was correct, but also said the answer with be a 4th degree

Answer 46

It is all confusing what you are calling problem2 and problem 3. I came in the middle and I saw two question that asks to apply Descartes rule of signs.

Answer 47

No, this has nothing to do with descartes rule of signs

Answer 48

The question is on it's own and I wanted to know if my function I gave was correct. I will be back in 20-30 min. Sorry

Answer 49

How did you arrive at your answer of x^3-9x^2+28x-30 ?

Answer 50

You answer x^3-9x^2+28x-30 does indeed have the roots: 3, 3 + i and 3 - i.

But the question has the real root 3 repeated twice (unless that was a typo).
So you need to multiply your answer by (x-3) to have the root 3 repeated twice.

For example, (x-2) has the root 2.
But (x-2)^2 has the root 2 twice. A second degree polynomial MUST have two roots and in this case both roots are 2.

Answer 51

The notifications and messaging system is not working properly for the past few days. So I will not know if you reply, tag or message me.

Answer 52

It was 3,3, and 3-i. What would be the function then?

Answer 53

@ranga maybe there is a delay

Answer 54

The function x^4-12x^3+55x^2-114x+90 still gives me the same 3,3+i, 3-i

Answer 55

Yes, except 3 is repeated twice and normally we wont say for example that the roots of (x-2)^2 is 2,2. We will say it just once. The root is 2.

x^4-12x^3+55x^2-114x+90 is a fourth degree polynomial. It must have 4 roots. Two of them are complex. So the remaining two roots must be real. But if we graph it we will notice it cuts/touches the x axis at only one point, namely, x = 3. That is because the root 3 has a multiplicity of two. The root occurs twice.

Answer 56

Ok I get that. However, what would be the final answer I would write

Answer 57

x^4-12x^3+55x^2-114x+90 is the polynomial they are asking for.

Answer 58

Ok if your 100% sure that's the answer then thank you! Your really a lot of help

Answer 59

you are welcome. If there was no typos in your question and they do want the polynomial to have the root 3 twice then that is the answer.

Answer 60

You are welcome.

If there was no typos in your question and they do want the polynomial to have the root 3 twice then the fourth degree polynomial is the answer.

If they want root 3 to occur only once then your cubic equation would be the answer.

Answer 61

Ok the teachers has 3 twice and no indication of a typo,so I'll go with the 4th degree polynomial