Find the equation of the tangent to the curve

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Find the equation of the tangent to the curve [tex]y=x^3-7x^2+14x-8[/tex] at the point where [tex]x = 1[/tex]. [tex]\text{Answer: }y = 3x -3[/tex]
Find the x-coordinate of the point at which the tangent is parallel to the tangent at [tex]x = 1[/tex].

I need help on the second part.
 

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  • #2
arildno
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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*
 
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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.
 
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arildno
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Well, do you agree that what you need to solve is the equation (written with "x"):
[tex]3=3x^{2}-14x+14[/tex]
This can be rewritten as:
[tex]3(x+1)(x-1)-14(x-1)=0\to(3(x+1)-14)(x-1)=0[/tex]
What must then x* be?
 
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Oh! Thank you!
 

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