Find the equation of the tangent to the curve

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The discussion focuses on finding the equation of the tangent to the curve y=x^3-7x^2+14x-8 at x=1, which is determined to be y=3x-3. Participants engage in solving for the x-coordinate where the tangent is parallel to this line, requiring the derivative at that point to equal 3. The derivative is established as y'(x)=3x^2-14x+14, leading to the equation 3=3x^2-14x+14. The equation is factored to find the values of x that satisfy it, with guidance provided on how to approach the solution. The conversation concludes with a sense of understanding on how to find the required x-coordinate.
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Find the equation of the tangent to the curve y=x^3-7x^2+14x-8 at the point where x = 1. \text{Answer: }y = 3x -3
Find the x-coordinate of the point at which the tangent is parallel to the tangent at x = 1.

I need help on the second part.
 
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Well, let x* be the point you seek.
What do you know of x*?
Do you agree that we must have y'(x*)=3?
That is, the derivatives of y(x) must be equal at x=1 and x=x*
 
arildno said:
Do you agree that we must have y'(x*)=3?
Yes.
arildno said:
That is, the derivatives of y(x) must be equal at x=1 and x=x*
I think I get it. But I still don't know how to get the answer.
 
Well, do you agree that what you need to solve is the equation (written with "x"):
3=3x^{2}-14x+14
This can be rewritten as:
3(x+1)(x-1)-14(x-1)=0\to(3(x+1)-14)(x-1)=0
What must then x* be?
 
Oh! Thank you!
 
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