Question

Find the slope of the tangent to the curve $y=x^3-x+1$ at the point whose x-coordinate is $1$. Also find the equation of normal at the same point.

Answer 1

$y=x^3-x+1$

${\frac{dy}{dx}}_{x=1}=3x^2-1=3-1=2$

Thus the slope of the tangent at $x=1$ is 2.

$\therefore$ slope of the normal at $x=1$ becomes $\frac{-1}{2}$

Thus, its equation can be written as $y=\frac{-x}{2}+c$

At $x=1,y=1-1+1=1$

Thus (1,1) lies on the normal and so, $1+\frac{1}{2}=c=\frac{3}{2}$

Thus, equation of the normal is $2y+x=3$