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^{3}-6X

^{2}+9x

and finf max and min turning points

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- Thread starter anthonyk2013
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- #1

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and finf max and min turning points

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SteamKing

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View attachment 77039 I'm wondering if I'm right or wrong. question is Apply differentiation to determine the co-ordinates of the turning points on the graph Y=X^{3}-6X^{2}+9x

and finf max and min turning points

When you are solving the equation x

This equation has two factors, namely (x - 3) and (x - 1). Re-writing x

- #3

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When you are solving the equation x^{2}- 4x + 3 = 0, you made a silly mistake.

This equation has two factors, namely (x - 3) and (x - 1). Re-writing x^{2}- 4x + 3 = (x - 3)(x - 1) = 0. By setting each factor equal to zero independently, you can make the equation true; thus x - 3 = 0 or x - 1 = 0. The solutions you have, x = -3 and x = -1, are incorrect.

Ok I did have trouble they're , i'm a long time away from this type of stuff and trying to remember from 25 years ago. Where should I go fro here so?

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SteamKing

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I used quadratic formula and got x=1,x=3

sub in x=3

Y=x^{3}-6x^{2}+9

Y=(3)^{3}-6(3)^{2}+9

Y=27-54+9

Y=-18

sub in x=1

Y=(1)^{3}-6(1)^{2}+9

Y=1-6+9

Y=4

dy/dx=3x^{2}-12x+9

d^{2}y/dx^{2}=6x-12

X=1, 6(1)-12=-6

X=3, 6(3)-12=6

sub in x=3

Y=x

Y=(3)

Y=27-54+9

Y=-18

sub in x=1

Y=(1)

Y=1-6+9

Y=4

dy/dx=3x

d

X=1, 6(1)-12=-6

X=3, 6(3)-12=6

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