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

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The discussion focuses on finding the derivative of the cubic function y = 2x^3 - x^2 + cx + d, given that the line 2y + 4x - 6 = 0 is tangent to the curve. The derivative is calculated as y' = -12x^2 + 2x - 2. By equating this derivative to the slope of the tangent line (-2), participants derive two x-values, which are then used to express c and d. The final values determined are c = -2 and d = 3, with the tangent point identified as (0, 3).

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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??
 

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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:
 

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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.
 
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Glad to help, I found the problem to be interesting, even if I did interpret it incorrectly! :)
 

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