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Find points where tangent line is horizontal

  • Thread starter Dustobusto
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  • #1
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Homework Statement



Find the points on the graph of x3-y3=3xy-3 where the tangent line is horizontal


Homework Equations



y = f(x) so implicit differentiation must be used when taking the derivative of y

(xy)' = xy' + y


The Attempt at a Solution



So if the tangent line is horizontal the slope = 0 and the derivative of y' must be equal to zero.

So first taking the derivative I end up with

3x2-3y2y' = 3xy' + 3y

Factor out the three and divide by three on both sides to cancel them out. That leaves me with

x2-y2y' = xy' + y

Subtract y on both sides, add y2y' on both sides and get

x2-y = xy' + y2y'

Factor out the y' to get

x2-y = y'(x+y2)

divide by x+y on both sides and

y' = (x2-y)/(x+y2)

So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?
 
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Answers and Replies

  • #2
818
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So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?
Remove it how?

You're looking for points (x,y) where dy/dx = 0. You have dy/dx = some fraction. When is a fraction zero?
 
  • #3
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Remove it how?

You're looking for points (x,y) where dy/dx = 0. You have dy/dx = some fraction. When is a fraction zero?
Well I understand part of what you're saying.

I can solve the y' so that x2 = y and then go back into the equation of y' and replace y with x2 and demonstrate that it equals zero. Do I plug x2 into the original y and solve for x? Or take the derivative and solve for x?
 
  • #4
Dick
Science Advisor
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Well I understand part of what you're saying.

I can solve the y' so that x2 = y and then go back into the equation of y' and replace y with x2 and demonstrate that it equals zero. Do I plug x2 into the original y and solve for x? Or take the derivative and solve for x?
Second choice. Substitute y=x^2 into the original equation and find x. Having a horizontal tangent at (x,y) means it has to satisfy the original equation AND y'=0. Two equations.
 
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  • #5
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OK, so x3-y3 = 3xy -3

Substitute,

x3 -(x2)3 = 3(x)(x2) - 3

x3 - x6 = 3x3-3

Subtract 3x3 both sides and get

-2x3-x6 = -3

So at this point, I'm not sure whats factorable and whats not. I could factor out an x3 and get

x3(-2-x3) = -3 but I'm not sure how much that helps me. Do I need to use the quadratic formula at some point?

NOPENOPENOPE I think I got it. I believe x = √3 cubed root rather
 
  • #6
ehild
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I think I got it. I believe x = √3 cubed root rather
Not quite. And what about the other root?

ehild
 
  • #7
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Not quite. And what about the other root?

ehild
Yeah I'm starting to realize that wasn't right

x3 - x6 = 3x3-3

is that at least on the right track?
 
  • #8
Dick
Science Advisor
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Yeah I'm starting to realize that wasn't right

x3 - x6 = 3x3-3

is that at least on the right track?
Fine so far. Move everything to one side and try to factor if. It's a quadratic in u=x^3.
 

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