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The table of values represents a polynomial function f (x).

How much greater is the average rate of change over the interval  [5, 7]  than the interval [2, 4]?

 

X values:      2       3      4     5      6       7

F(x) values: 39   125  287 549 935 1469

 

Thanks for all of your help

Guest Mar 30, 2018

Best Answer 

 #1
avatar+6943 
+3

avg rate of change over the interval [5, 7]   =   \(\frac{f(7)-f(5)}{7-5}\)

 

avg rate of change over the interval [5, 7]   =   \(\frac{1469-549}{7-5}\)

 

avg rate of change over the interval [5, 7]   =   \(\frac{920}{2}\)

 

avg rate of change over the interval [5, 7]   =   460

 

avg rate of change over the interval [2, 4]   =   \(\frac{f(4)-f(2)}{4-2}\)

 

avg rate of change over the interval [2, 4]   =   \(\frac{287-39}{4-2}\)

 

avg rate of change over the interval [2, 4]   =   \(\frac{248}{2}\)

 

avg rate of change over the interval [2, 4]   =   124

 

How much greater is the avg rate of change over the interval  [5, 7]  than the avg rate of change over the interval  [2, 4] ? That is...how much greater is  460  than  124 ?

 

460 - 124  =  336

 

460  is  336  greater than  124 .

hectictar  Mar 30, 2018
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1+0 Answers

 #1
avatar+6943 
+3
Best Answer

avg rate of change over the interval [5, 7]   =   \(\frac{f(7)-f(5)}{7-5}\)

 

avg rate of change over the interval [5, 7]   =   \(\frac{1469-549}{7-5}\)

 

avg rate of change over the interval [5, 7]   =   \(\frac{920}{2}\)

 

avg rate of change over the interval [5, 7]   =   460

 

avg rate of change over the interval [2, 4]   =   \(\frac{f(4)-f(2)}{4-2}\)

 

avg rate of change over the interval [2, 4]   =   \(\frac{287-39}{4-2}\)

 

avg rate of change over the interval [2, 4]   =   \(\frac{248}{2}\)

 

avg rate of change over the interval [2, 4]   =   124

 

How much greater is the avg rate of change over the interval  [5, 7]  than the avg rate of change over the interval  [2, 4] ? That is...how much greater is  460  than  124 ?

 

460 - 124  =  336

 

460  is  336  greater than  124 .

hectictar  Mar 30, 2018

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