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

How much greater is the average rate of change over the interval [7, 9] than the interval [4, 6]?



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Sagot :

MrRoyalidn

Answer:

It is 603 units greater

Step-by-step explanation:

Given

See attachment for table

Average rate of change over (a,b) is calculated as:

[tex]Rate = \frac{f(b) - f(a)}{b-a}[/tex]

For interval [7,9], we have:

[tex][a,b] = [7,9][/tex]

So, we have:

[tex]Rate = \frac{f(9) - f(7)}{9-7}[/tex]

[tex]Rate = \frac{f(9) - f(7)}{2}[/tex]

From the table:

[tex]f(9) = 3878[/tex]

[tex]f(7) = 1852[/tex]

So:

[tex]Rate = \frac{f(9) - f(7)}{2}[/tex]

[tex]Rate = \frac{3878 - 1852}{2}[/tex]

[tex]Rate = \frac{2026}{2}[/tex]

[tex]Rate = 1013\\[/tex]

For interval [4,6], we have:

[tex][a,b] = [4,6][/tex]

So, we have:

[tex]Rate = \frac{f(6) - f(4)}{6-4}[/tex]

[tex]Rate = \frac{f(6) - f(4)}{2}[/tex]

From the table:

[tex]f(6) = 1178[/tex]

[tex]f(4) = 358[/tex]

So:

[tex]Rate = \frac{f(6) - f(4)}{2}[/tex]

[tex]Rate = \frac{1178 - 358}{2}[/tex]

[tex]Rate = \frac{820}{2}[/tex]

[tex]Rate = 410[/tex]

Calculate the difference (d) to get how much greater their rate of change is:

[tex]d = 1013 - 410[/tex]

[tex]d = 603[/tex]

View image MrRoyal
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