Question

F(x)=x^3-9 What is the average rate of change of f over the interval [1,6]?

Answer 1

SOLUTION

GIVEN

$\sf F(x) = {x}^{3} - 9$

TO DETERMINE

The average rate of change of F over the interval [1,6]

FORMULA TO BE IMPLEMENTED

For a given function F(x) , the average rate of change of f over the interval [a, b]

$\displaystyle \sf{ = \frac{F(b) - F(a)}{b - a} }$

EVALUATION

Here the given function is

$\sf F(x) = {x}^{3} - 9$

The given interval is [1,6]

Now

$\sf F(6) = {6}^{3} - 9 = 216 - 9 = 207$

$\sf F(1) = {1}^{3} - 9 = 1 - 9 = - 8$

Now the average rate of change of F over the interval [1,6]

$\displaystyle \sf{ = \frac{F(6) - F(1)}{6 - 1} }$

$\displaystyle \sf{ = \frac{207 - ( - 8)}{5} }$

$\displaystyle \sf{ = \frac{207 + 8}{5} }$

$\displaystyle \sf{ = \frac{215}{5} }$

$\displaystyle \sf{ = 43}$

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Answer 2

Answer:

34

Step-by-step explanation:

ARC [1,6] = f(6) - f(1) / 6-1

f(1)= 1^3 -9*1 = 8

f(6)= 6^3 -9*6 = 162

162-(-8)/6-1 = 170/5

170/5 = 34

Average rate of change of f over the interval [1,6] is 34