Question

Find the slope of the tangent and the normal to the following curve at the indicted point  y = x³ − x at x = 2 ?

Answer 1

\[y = x^3 - x\]

\[ \Rightarrow \frac{dy}{dx} = 3 x^2 - 1\]

When `x=2,`

`y=x^3-x`

`=2^3-2`

`=8-2`

`=6`

\[\text { Now }, \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 2, 6 \right) =3 \left( 2 \right)^2 -1=11\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 2, 6 \right)}=\frac{- 1}{11}\]