+4622 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
+4622 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.
+1995 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?
+54 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.
+54 Yes, you stop when there are all positive values. Feel free to message me at any time!
+1995 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!
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
+4622 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.
+1995 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?
+54 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.
+54 Yes, you stop when there are all positive values. Feel free to message me at any time!
+1995 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!
+129898 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
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3 -1 we get a negative so we can stop
Test 3
3 [ 3 - 7 0 5 -3]
9 6 18 69
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3 2 6 23 66 we ge all positives....so 3 is the largest integer upper bound
+1995 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!
+129898 f(x) = 4x^4 - 25x^2 -5x - 13
Test 1 using synthetic division
1 [ 4 0 -25 -5 -13 ]
4 4
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4 4 -21 stop
Test 2
2 [ 4 0 -25 - 5 -13 ]
8 16
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4 8 -9 stop
Test 3
3 [ 4 0 -25 -5 -13 ]
12 36 33 84
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4 12 11 28 71 all are positive......3 is the upper bound
