+0  
 
0
324
16
avatar+1898 

 

i think its 3 ?

 Sep 14, 2019
 #1
avatar+790 
+2

i believe it is -3

 Sep 14, 2019
 #2
avatar+4593 
+2

We try the values y= 1, 2, 3, 4, 5, 6,...and so on.

 

For the function to have an upper bound, the coefficients of the polynomial when the original polynomial divided by (y) has to be positive. I recommend using Sythetic Division to divide the polynomial with the number.

 

Trying y=1, gives us a polynomial that alternates in signs. Same with y=2. Yet, y=3 gives us a polynomial with all positive coefficients. y=3: Answer

 Sep 14, 2019
edited by tertre  Sep 14, 2019
edited by tertre  Sep 14, 2019
 #5
avatar+4593 
+2

Posted two responses at once.

tertre  Sep 14, 2019
 #7
avatar+1898 
0

 

is this one correct as well ?

jjennylove  Sep 15, 2019
 #8
avatar+1898 
0

thank you for checking my answer and taking the time to help smiley

jjennylove  Sep 15, 2019
 #10
avatar+4593 
0

Testing the values with y=1, y=2, y=3, y=4 yields negative value by using Synthetic Division. Yet, we can realize something. The -25x^3 is really messing up the positive values, so we need to find a value larger than 25. Since 6*4=24, that won't work, yet y=7 will work since 7*4=28, which is greater than 25.And yes, y=7 does work.

tertre  Sep 15, 2019
 #11
avatar+1898 
+1

When i solve 6 for 4 0 -25 -5 -13 i as well get all postitive vaules, dnt i need to find the lowest value which gives all postitive or all negatitive values?

jjennylove  Sep 15, 2019
 #12
avatar+54 
+1

You're correct, six does work. 

neworleans06  Sep 15, 2019
 #13
avatar+54 
+1

Tertre might have typoed, yet I think he forgot the 0x^2. This is really important! Check three just to make sure. Three should work.

neworleans06  Sep 15, 2019
 #14
avatar+54 
+1

Yes, you stop when there are all positive values. Feel free to message me at any time!

neworleans06  Sep 15, 2019
 #15
avatar+1898 
0

i jsut checked three as well and it worked , the answer must be 3 right? since its the lowest value before they all turn into a mix of pos and neg values? Thank you so much for your help!

jjennylove  Sep 15, 2019
edited by jjennylove  Sep 15, 2019
 #2
avatar
0

We try the values y= 1, 2, 3, 4, 5, 6,...and so on.

 

For the function to have an upper bound, the coefficients of the polynomial when the original polynomial divided by (y) has to be positive. I recommend using Sythetic Division to divide the polynomial with the number.

 

Trying y=1, gives us a polynomial that alternates in signs. Same with y=2. Yet, y=3 gives us a polynomial with all positive coefficients. y=3: Answer

Guest Sep 14, 2019
 #5
avatar+4593 
+2

Posted two responses at once.

tertre  Sep 14, 2019
 #7
avatar+1898 
0

 

is this one correct as well ?

jjennylove  Sep 15, 2019
 #8
avatar+1898 
0

thank you for checking my answer and taking the time to help smiley

jjennylove  Sep 15, 2019
 #10
avatar+4593 
0

Testing the values with y=1, y=2, y=3, y=4 yields negative value by using Synthetic Division. Yet, we can realize something. The -25x^3 is really messing up the positive values, so we need to find a value larger than 25. Since 6*4=24, that won't work, yet y=7 will work since 7*4=28, which is greater than 25.And yes, y=7 does work.

tertre  Sep 15, 2019
 #11
avatar+1898 
+1

When i solve 6 for 4 0 -25 -5 -13 i as well get all postitive vaules, dnt i need to find the lowest value which gives all postitive or all negatitive values?

jjennylove  Sep 15, 2019
 #12
avatar+54 
+1

You're correct, six does work. 

neworleans06  Sep 15, 2019
 #13
avatar+54 
+1

Tertre might have typoed, yet I think he forgot the 0x^2. This is really important! Check three just to make sure. Three should work.

neworleans06  Sep 15, 2019
 #14
avatar+54 
+1

Yes, you stop when there are all positive values. Feel free to message me at any time!

neworleans06  Sep 15, 2019
 #15
avatar+1898 
0

i jsut checked three as well and it worked , the answer must be 3 right? since its the lowest value before they all turn into a mix of pos and neg values? Thank you so much for your help!

jjennylove  Sep 15, 2019
edited by jjennylove  Sep 15, 2019
 #4
avatar+111438 
+2

Divide each term by 3

 

1, -7/3, 5/3, -1

 

Remove the first term    and change any minuses to pluses

 

7/3, 5/3/ 1

 

1) take the largest value and add 1 to it  →  7/3 + 1  = 10/3

 

2) Sum the values =  5   .....compare this with 1 and take the larger value  = 5

 

Take the smaller  of  (1), (2)

 

The upper bound is 10/3

 

See this graph  : https://www.desmos.com/calculator/1mf8fqbard

 

However....I  think that you are using the syntethic division test to find the greatest integer that is an upper bound.....this should be 3

 

Test 2

 

2 [  3  -7   0  5  -3]

           6

  _______________

     3  -1                              we get a negative so we can stop

 

Test  3

 

3 [  3 - 7   0  5   -3]

           9   6 18   69

   _______________

     3   2    6  23  66       we ge all positives....so 3 is the largest integer upper bound 

 

 

cool cool cool

 Sep 14, 2019
 #9
avatar+1898 
0

im not sure if you remember but this is the way  i was talking about for the other question posted above, because i didnt feel like 7.25 fit properly , do u think maybe the answer would then be 13? Thank you as well for your help!

jjennylove  Sep 15, 2019
 #6
avatar+111438 
0

Thanks, tertre....!!!

 

Good to see you ......!!!

 

cool cool cool

 Sep 14, 2019
 #16
avatar+111438 
+2

f(x)  = 4x^4 - 25x^2 -5x - 13

 

Test 1  using synthetic division

 

1  [ 4    0    -25     -5        -13 ]

            4       4

    ______________________

      4    4   -21       stop

 

Test 2

 

2 [  4   0     -25       - 5    -13  ]

           8     16

    ________________________

      4   8      -9      stop

 

Test 3

 

3  [ 4     0       -25       -5       -13  ]

             12      36      33        84

   ___________________________

     4     12      11      28        71            all are positive......3   is the upper bound

 

 

 

cool cool cool

 Sep 15, 2019

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