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

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Homework Help Overview

The discussion revolves around determining the coordinates of the turning points for the function Y = X³ - 6X² + 9X using differentiation. Participants are exploring the application of calculus concepts to identify maximum and minimum turning points.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply differentiation to find turning points and are questioning the correctness of previously derived solutions. There is a focus on solving a quadratic equation related to the function's critical points.

Discussion Status

Some participants have provided guidance on correcting earlier mistakes in solving the quadratic equation, while others are sharing their calculations for the turning points. Multiple interpretations of the problem are being explored, particularly regarding the differentiation process and the identification of turning points.

Contextual Notes

There is mention of a long gap since some participants last engaged with similar mathematical concepts, indicating potential challenges in recalling methods. The discussion also highlights specific solutions that were deemed incorrect, prompting further examination of the problem.

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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
 
Last edited:

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