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footprints
Mar23-05, 10:45 AM
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.

arildno
Mar23-05, 10:49 AM
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*

footprints
Mar23-05, 10:58 AM
Do you agree that we must have y'(x*)=3?
Yes.
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.

arildno
Mar23-05, 11:07 AM
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?

footprints
Mar23-05, 11:15 AM
Oh! Thank you!