Find the equation of the line that is tangent to the curve y=x^3 and is parallel to the line 3x−y+1=0


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Question Number 31812 by NECx last updated on 15/Mar/18

$F i n d t h e e q u a t i o n o f t h e l i n e$ $t h a t i s t a n g e n t t o t h e c u r v e y = x^{3}$ $a n d i s p a r a l l e l t o t h e l i n e$ $3 x - y + 1 = 0$
	Answered by mrW2 last updated on 15/Mar/18

	$l e t y = 3 x + c b e t h e l i n e t a n g e n t t o y = x^{3}$ $\frac{d y}{d x} = 3 x^{2} = 3$ $\Rightarrow x = \pm 1$ $y = 3 x + c = \pm 3 + c = \pm 1$ $\Rightarrow c = - 2 o r 2$ $\Rightarrow e q n . o f l i n e i s$ $y = 3 x - 2 o r$ $y = 3 x + 2$
	Answered by ajfour last updated on 15/Mar/18

	$l e t e q . o f t a n g e n t t o y = x^{3} a t (x_{1}, x_{1}^{3}) b e$ $y - x_{1}^{3} = 3 x_{1}^{2} (x - x_{1})$ $a s \frac{d y}{d x} ∣_{x = x_{1}} = 3 (s l o p e o f g i v e n / / l i n e)$ $\Rightarrow 3 x_{1}^{2} = 3 \Rightarrow x_{1} = \pm 1$ $s o r e q d . t a n g e n t e q s . a r e$ $y - 1 = 3 (x - 1) o r y = 3 x - 2$ $a n d y + 1 = 3 (x + 1) o r y = 3 x + 2 .$
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