Solving for Derivative of 2x^3-x^2+cx+d

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Discussion Overview

The discussion revolves around finding the derivative of the cubic function \(y = 2x^3 - x^2 + cx + d\) at a point where the curve is tangent to the line given by the equation \(2y + 4x - 6 = 0\). Participants explore the conditions necessary for tangency and the implications for the coefficients \(c\) and \(d\).

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant asks for help in solving the problem of finding the derivative at a point of tangency with the given line.
  • Another participant questions what conditions are needed for the line to be tangent to the curve.
  • It is noted that the gradient of the line is \(-2\), and a participant expresses uncertainty about how to find the coefficients \(c\) and \(d\).
  • Several participants suggest equating the derivative of the cubic function to \(-2\) to express \(c\) as a function of \(x\) and find the corresponding \(x\)-values.
  • One participant confirms the derivative found is \(y' = -12x^2 + 2x - 2\) and discusses the process of finding \(c\) and \(d\) through substitution.
  • Another participant mentions that the exercise is conceptual, emphasizing the relationship between the derivative and the gradient of the tangent line.
  • Some participants assume specific values for \(c\) and \(d\) based on a presumed tangent point, suggesting \(c = -2\) and \(d = 3\).
  • A participant shares a plot related to the discussion, although the relevance of the plot is not clarified.

Areas of Agreement / Disagreement

Participants generally agree on the method of equating the derivative to find the coefficients, but there is no consensus on the specific values of \(c\) and \(d\) or the exact point of tangency. Some assumptions about these values are made, but they are not universally accepted.

Contextual Notes

There are unresolved assumptions regarding the specific point of tangency and the values of \(c\) and \(d\). The discussion does not clarify the dependence on the chosen point or the implications of different potential tangent points.

Chipset3600
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Hello MHB, please help me solve this problemn.

knowing that 2y+4x-6=0 is the equation which is one of the lines tangent to the curve y= 2x^3-x^2+cx+d.
Find the derivative of this function in one of the points of the curve.
 
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What do you require of the derivative and the curve so that the given line is tangent to the curve?
 
MarkFL said:
What do you require of the derivative and the curve so that the given line is tangent to the curve?
this implicit function "2y+4x-6=0" is tangent of this curve "y= 2x^3-x^2+cx+d"
the gradient is "-2", but i don't know my C and D. How can i find the common point?
 
Equate the derivative of the cubic to -2, and you may express c as a function of x.

Then, substitute for c into the cubic and find where this new function's derivative is equal to -2. You will get 2 x-values. One of these will allow you to then easily find c and d.
 
MarkFL said:
Equate the derivative of the cubic to -2, and you may express c as a function of x.

Then, substitute for c into the cubic and find where this new function's derivative is equal to -2. You will get 2 x-values. One of these will allow you to then easily find c and d.

Lets see if i understood: View attachment 453

Is this what you mean??
 

Attachments

  • load.jpg
    load.jpg
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Yes, after substituting for c, then taking the derivative, you have found:

$\displaystyle y'=-12x^2+2x-2$

Now, equate this to -2, and you will have 2 roots, one of which allows you to find by substitution into the original cubic and tangent line the value of d (when you equate the two as they must have the same value where they touch) and then use the x-value also in your expression for c to find its value.
 
MarkFL said:
Yes, after substituting for c, then taking the derivative, you have found:

$\displaystyle y'=-12x^2+2x-2$

Now, equate this to -2, and you will have 2 roots, one of which allows you to find by substitution into the original cubic and tangent line the value of d (when you equate the two as they must have the same value where they touch) and then use the x-value also in your expression for c to find its value.

My teacher said that this is just an exercise concept.
knowing that the derivative in point is the gradient of the tangent line
And he give the equation("2y+4x-6=0") of one of tangent line in some point.
The answer is: m(x)= -2...
 
I assumed your teacher wanted you to find a tangent point (0,3), and the values of c and d, which are -2 and 3 respectively.

Here is a plot:
 

Attachments

  • cubicandtangent.jpg
    cubicandtangent.jpg
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MarkFL said:
I assumed your teacher wanted you to find a tangent point (0,3), and the values of c and d, which are -2 and 3 respectively.

Here is a plot:

Every body thought that, thanks for the help MarkFL.
 
  • #10
Glad to help, I found the problem to be interesting, even if I did interpret it incorrectly! :)
 

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