So the coordinates of the turning points are (1,4) and (3,-18). Is this correct?

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The discussion focuses on determining the coordinates of turning points for the function Y=X^3-6X^2+9X using differentiation. The correct turning points are identified as (1, 4) and (3, -18) after applying the quadratic formula to the derivative. A mistake was noted in solving the equation x^2 - 4x + 3 = 0, where incorrect solutions were initially provided. Participants emphasized the importance of verifying calculations and using correct solutions to proceed accurately. The final results confirm the turning points and the process of differentiation used to find them.
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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=X3-6X2+9x
and finf max and min turning points
 
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anthonyk2013 said:
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=X3-6X2+9x
and finf max and min turning points

When you are solving the equation x2 - 4x + 3 = 0, you made a silly mistake.

This equation has two factors, namely (x - 3) and (x - 1). Re-writing x2 - 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.
 
SteamKing said:
When you are solving the equation x2 - 4x + 3 = 0, you made a silly mistake.

This equation has two factors, namely (x - 3) and (x - 1). Re-writing x2 - 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?
 
Correct the solutions you have and work from there. All of your work past this point must be checked again using the correct solutions.
 
I used quadratic formula and got x=1,x=3
sub in x=3
Y=x3-6x2+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=3x2-12x+9

d2y/dx2=6x-12

X=1, 6(1)-12=-6
X=3, 6(3)-12=6
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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