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# stationary point rational function

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4 ok so for this question i have showed one of the coordinate is (1,0) what im struggling with is the other stationary point.

i simplified the rational function into (3x^2 - 6x +3)/(2x^2 + 6)

then i differentiated the numerator into 6x-6

using the quotent rule i have

(6x-6)*(2x^2+6) - (3x^2 - 6x +30) (6x -6)

i expanded and simplified them into

-6x^3 + 12x^2 - 18x -18

what should i do from here?

Dec 29, 2018

#1
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We don't need to worry about the denominator with the quotient rule

Remember that the quotient rule is  ....   (f / g)'  =   [f' g -   f g'] / [ g^2 ]

f = 3(x - 1)^2  =  3x^2 - 6x + 3

g = 2x^2 + 6

So....

The numerator is

(6x - 6)(2x^2 + 6)  -  (3x^2 - 6x + 3) (4x)   =  0

6 ( x - 1) * 2 (x^2 + 3)  -  3(x - 1)^2(4x) = 0

12(x - 1) (x^2 + 3) - 12x(x - 1)^2  =  0

12(x - 1)  [ (x^2 + 3) - (  x(x - 1) ]  = 0

12 (x - 1) [ x^2 + 3 - x^2 + x]  = 0

12(x - 1) (x + 3 )  = 0

Setting both factors to 0 and solving for x produces the two stationary x coordinates of  x = 1    and x = - 3

And f(-3) = 2

So....the other stationary point is (-3, 2)   Dec 29, 2018
edited by CPhill  Dec 29, 2018
#2
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Can you explain where the 4x came from please thanks very much

YEEEEEET  Dec 29, 2018
#3
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It's the derivative of  2x^2 + 6    =   4x   CPhill  Dec 29, 2018
#4
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ok thank you

YEEEEEET  Dec 29, 2018