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Old 10-24-2010   #1
magmagod
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Smile Rates of change question

Use an algebric strategy to verify that the point given for each function is either a maximum or a minimum.

x^3-3x ; (-1, 2)

so i tried finding the instantaneous rate of change before -1 and after -1 to see if its a maximum or a minimum, since if its a local max before -1 would be positive and after would be negative but i couldn't reach the solution in the back of the book. any help would be appreciated.

the formula i used was the difference quotient.
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Old 11-03-2010   #2
magmagod
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Originally Posted by magmagod View Post
Use an algebric strategy to verify that the point given for each function is either a maximum or a minimum.

x^3-3x ; (-1, 2)

so i tried finding the instantaneous rate of change before -1 and after -1 to see if its a maximum or a minimum, since if its a local max before -1 would be positive and after would be negative but i couldn't reach the solution in the back of the book. any help would be appreciated.

the formula i used was the difference quotient.
never mind, i got the answer..
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Old 01-06-2011   #3
Jonathan W
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Originally Posted by magmagod View Post
never mind, i got the answer..
I got the answer too but I forgot
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Old 01-07-2011   #4
Lisasmith111
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Originally Posted by magmagod View Post
Use an algebric strategy to verify that the point given for each function is either a maximum or a minimum.

x^3-3x ; (-1, 2)

so i tried finding the instantaneous rate of change before -1 and after -1 to see if its a maximum or a minimum, since if its a local max before -1 would be positive and after would be negative but i couldn't reach the solution in the back of the book. any help would be appreciated.

the formula i used was the difference quotient.
Well I thought I'd have a go at this one anyway!

Here are my thoughts:-

f(x) = x^3 - 3x; verify point (-1,2)

f '(x) = 3x^2 -3

For max or min f '(x) = 0 so we get:-

3x^2 -3 = 0 (divide this though by 3 we get next line)

x^2 -1 = 0

then factorising we get

(x + 1) (x - 1) = 0

so x = -1 or x = 1

Substituting this back into the first eqn f (x) we can get the Y coordinates and determine whether local max or min.

so f (-1) = (-1)^3 -3(-1) = -1 + 3 = 2 so coordinates are (-1, 2) = Y max

and f (1) = (1)^3 -3(1) = 1 -3 = -2 so coordinates are (1 , -2) = Y min

Remember these are local max and mins not the highest or lowest points on the curve of f (x) = x^3 - 3x.

See diagram of curve from x = -3 to x = 3 for confirmation.

Hope this is of use for someone
Attached Images
File Type: bmp Y = x^3 -3x.bmp (850.3 KB, 20 views)
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