SOLUTION: Find the average rate of change of f (x)= =-3x^2 from x=—3 to x=-1.

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Question 1189588: Find the average rate of change of f (x)= =-3x^2 from x=—3 to x=-1.
Found 3 solutions by Alan3354, Theo, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find the average rate of change of f (x)= =-3x^2 from x=—3 to x=-1.
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f(-3) = -27
f(-1) = -4
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Change of 23/2 = 11.5
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the equation is f(x) = -3 * x^2

the value of x is from -3 to -1.

when x = -3, f(x) = -3 * (-3)^2 = -3 * 9 = -27
when x = -1, f(x) = -3 * (-1)^2 = -3 * 1 = -3

you can use the geometric series formula to solve this.

the geometric series formula is An = A1 * r^(n-1).

A1 = the value of f(x) when x = -3.
that becomes -3 * (-3)^2 = -3 * 9 = -27

the formula becomes An = -27 * r^(n-1)

A3 is equal to f(x) when x = -1 because x = -1 is 2 units more positive than x = -3.

the formula becomes A3 = -27 * r^(3-1) which becomes A3 = -27 * r^2.
when An = A3, the value of n is equal to 3.
the value of A3 is the value of f(x), when x = -1.
that becomes -3 * (-1)^2 = -3.

the formula becomes -3 = -27 * r^(3-1) which becomes -3 = -27 * r^2.

solve for r to get:

r = (-3/-27)^(1/2) = .333333333.....

what this says is that, each successive value in the geometric series is 1/3 the value of the preceding value.

you start at A1 = -27
then A2 = -9
then A3 = -3.

your common ratio is .3333333..... = 1/3.

that's your average rate of change.
each succeeding term in the sequence is 1/3 * the value of the the preceding term.



Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
Find the average rate of change of f (x)= =-3x^2 from x= -3 to x= -1.
~~~~~~~~~~~~~~~ 

    the average rate of change = BY THE DEFINITION = .    (*)


f(-1) =  = -3 * 1 = -3;

f(-3) =  = -3 * 9 = -27.


THEREFORE

    the average rate of change =  =  = 12.    ANSWER

Solved, answered and explained.


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Ignore the solutions and the answers of both preceding tutors' posts, since they are INCORRECT.

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