Intermediate Value Theorem

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Section 2.2
Question 90

Can you do 90 (a - b) as a guide for me to do 91 (a - b) on my own?

What is the Intermediate Value Theorem?

20210827_203714.jpg
 
90.
use intermediate value theorem and the table to find intervals 1 unit in the length in which the polynomial is quarantined to have a zero

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis.

upload_2021-8-30_13-55-52.png




Let f be a polynomial function. The Intermediate Value Theorem states that if f(a) and f(b) have opposite signs, then there exists at least one value c between a and b for which f(c)=0.

As a start, evaluate f(x) at the integer values x=0,1,2,3,4,......,

f(x)=0.11x^3-2.07x^2+9.81x-6.88
first graph
MSP293214e2926280702gb400003g83d498156h0db7



table ( choosing x observing the graph)

x| f(x)
0| -6.88
1| 0.97.......................f(1)=0.11(1)^3-2.07(1)^2+9.81(1)-6.88=0.97
6| 1.22.......................f(6)=0.11(6)^3-2.07(6)^2+9.81(6)-6.88=1.22
7| -1.91.......................f(7)=0.11(7)^3-2.07(7)^2+9.81(7)-6.88=-1.91
11| -3.03.......................f(11)=0.11(11)^3-2.07(11)^2+9.81(11)-6.88=-3.03
12| 2.84.......................f(12)=0.11(12)^3-2.07(12)^2+9.81(12)-6.88=2.84


[0,1]->This interval does work because as x increases, f(x) changes signs. There is a zero.

[6,7]->This interval does work because as x increases, f(x) changes signs. There is a zero.
[11,12]->This interval does work because as x increases, f(x) changes signs. There is a zero.

We have shown that there are at least two real zeros between x=0 and x=1, x=6 and x=7, x=11 and x=12

b.
adjust the table to approximate the zeros of the function to the nearest thousandth

let's check x=0.839 , x=6.345, and x=11.612
x|f(x)
0.839| -0.042.......................f(0.839)=0.11(0.839)^3-2.07(0.839)^2+9.81(0.839)-6.88=-0.301->round to 0
6.345| 0.269.......................f(6.345)=0.11(6.345)^3-2.07(6.345)^2+9.81(6.345)-6.88=0.127->round to 0
11.612| 0.18.......................f(11.612)=0.149->round to 0

approximate zeros are: x=0.839 , x=6.345, and x=11.612

find actual zeros:
easiest way is to factor
f(x)=0.11x^3-2.07x^2+9.81x-6.88....... factor completely
f(x) = 0.11 (x - 11.5878) (x - 6.38506) (x - 0.845338)

zeros are at:

x≈0.845
x≈6.385
x≈11.588
 
Last edited:
90.
use intermediate value theorem and the table to find intervals 1 unit in the length in which the polynomial is quarantined to have a zero

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis.

View attachment 353



Let f be a polynomial function. The Intermediate Value Theorem states that if f(a) and f(b) have opposite signs, then there exists at least one value c between a and b for which f(c)=0.

As a start, evaluate f(x) at the integer values x=0,1,2,3,4,......,

f(x)=0.11x^3-2.07x^2+9.81x-6.88
first graph
MSP293214e2926280702gb400003g83d498156h0db7



table ( choosing x observing the graph)

x| f(x)
0| -6.88
1| 0.97.......................f(1)=0.11(1)^3-2.07(1)^2+9.81(1)-6.88=0.97
6| 1.22.......................f(6)=0.11(6)^3-2.07(6)^2+9.81(6)-6.88=1.22
7| -1.91.......................f(7)=0.11(7)^3-2.07(7)^2+9.81(7)-6.88=-1.91
11| -3.03.......................f(11)=0.11(11)^3-2.07(11)^2+9.81(11)-6.88=-3.03
12| 2.84.......................f(12)=0.11(12)^3-2.07(12)^2+9.81(12)-6.88=2.84


[0,1]->This interval does work because as x increases, f(x) changes signs. There is a zero.

[6,7]->This interval does work because as x increases, f(x) changes signs. There is a zero.
[11,12]->This interval does work because as x increases, f(x) changes signs. There is a zero.

We have shown that there are at least two real zeros between x=0 and x=1, x=6 and x=7, x=11 and x=12

b.
adjust the table to approximate the zeros of the function to the nearest thousandth

let's check x=0.839 , x=6.345, and x=11.612
x|f(x)
0.839| -0.042.......................f(0.839)=0.11(0.839)^3-2.07(0.839)^2+9.81(0.839)-6.88=-0.301->round to 0
6.345| 0.269.......................f(6.345)=0.11(6.345)^3-2.07(6.345)^2+9.81(6.345)-6.88=0.127->round to 0
11.612| 0.18.......................f(11.612)=0.149->round to 0

approximate zeros are: x=0.839 , x=6.345, and x=11.612

find actual zeros:
easiest way is to factor
f(x)=0.11x^3-2.07x^2+9.81x-6.88....... factor completely
f(x) = 0.11 (x - 11.5878) (x - 6.38506) (x - 0.845338)

zeros are at:

x≈0.845
x≈6.385
x≈11.588

I don't understand what you did for the table in part (b). In fact, both tables are hard to read with numbers scattered. Can you ckearly write each table by hand and post for me to read?
 
Section 2.2
Question 90

Can you do 90 (a - b) as a guide for me to do 91 (a - b) on my own?

What is the Intermediate Value Theorem?

View attachment 351
I don't understand what you did for the table in part (b). In fact, both tables are hard to read with numbers scattered. Can you ckearly write each table by hand and post for me to read?

yes, I did tables by my self
in this post there are no tables at all
I was searching for this problem and found these tables:


upload_2021-8-30_18-24-23.png


Adjust the table see the zeros of the function.

upload_2021-8-30_18-25-24.png



In the above display value is changing from negative to positive in the interval [0,1], therefore zero must be there in this interval.

upload_2021-8-30_18-26-23.png


In the above display value is changing from positive to negative in the interval [6,7], therefore zero must be there in this interval.
 
yes, I did tables by my self
in this post there are no tables at all
I was searching for this problem and found these tables:


View attachment 354

Adjust the table see the zeros of the function.

View attachment 355


In the above display value is changing from negative to positive in the interval [0,1], therefore zero must be there in this interval.

View attachment 356

In the above display value is changing from positive to negative in the interval [6,7], therefore zero must be there in this interval.

I plan to do at least two more on my own. My working be posted here.

1. Did you check out my word problems in the Basic Math forum?

2. Can you find me a video clip that explain the Intermediate Value Theorem?

I somewhat understand what you did here but there are certain topics that I understand better watching a video clip.
 
1. what is the link to Basic Math forum?

2.
just think about the value: if it is changing from negative to positive, means must cross x-axis (means there is a zero)

 
1. what is the link to Basic Math forum?

2.
just think about the value: if it is changing from negative to positive, means must cross x-axis (means there is a zero)


You are great. Thank you. The Basic Math forum is here. Go to the list of forums and click on Basic Math.
 
You are great. Thank you. The Basic Math forum is here. Go to the list of forums and click on Basic Math.

If the value changes from negative to positive and/or vice-versa there is a zero, which means the graph crosses the line x = 0.

Yes?
 
1. what is the link to Basic Math forum?

2.
just think about the value: if it is changing from negative to positive, means must cross x-axis (means there is a zero)


Look at the screenshot of thie main page. Do you see Basic Math? Go there to see all my word problems.

Screenshot_20210830-221535_Samsung Internet.jpg
 


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