Question

Find the points at which the tangents to the curves $y=f\left(x\right)=x^3-x-1andy=\phi \left(x\right)=3x^2-4x+1$ are parallel. Write the equation of the tangents.

Answer 1

$y=f\left(x\right)=x^3-x-1\int \left(x\right)=3x^2-1$

$y=\Phi \left(x\right)=3x^2-4x+1\Phi \left(x\right)=6x-4$

For parallel tangent, slope have to be equal.

$\therefore f^{\prime }\left(x\right)={\Phi }^{\prime }\left(x\right)\Rightarrow 3x^2-1=6x-4\Rightarrow 3x^2-6x+3=0$

$\Rightarrow 3x^2-3x-3x+3=0$

$\Rightarrow (x-1)=0$

$\Rightarrow x=1$

$f^{\prime }\left(1\right)=2$

$f\left(1\right)=-1$ and $\Phi \left(1\right)=0$

Tangent an $f\left(x\right)=at(1,-1)\Rightarrow \frac{y+1}{x-1}=2\Rightarrow y+2=2x-2$

$\Rightarrow 2x-y-4=0$

${\Phi }^{\prime }\left(1\right)=2$

Tangent an $\Phi \left(x\right)$ at $(1,0)\Rightarrow \frac{y-0}{x-1}=2\Rightarrow 2x-y-2=0$