Find the points on the curve `x^3-2x^2-2x` at which the tangent lines are parallel to the line `y=2x-3` .

Answer

`(x_{1}, y_{1})=(frac{-2}{3}, frac{4}{27})`

Explanation

Given:
`y=2 x-3`

`\therefore` Slope of the line `=\frac{d y}{d x}=2`


`y=x^{3}-2 x^{2}-2 x`

Since `(x_{1} y_{1})` lies on curve,`y_{1}=x_{1}^{3}-2 x_{1}^{2}-2 x_{1} \ldots(1)`

`\Rightarrow(\frac{d y}{d x})_{(x_{1}, y_{1}}=3 x_{1}^{2}-4 x_{1}-2`


As given tangent and line are parallel.

`\therefore` Slope of the tangent `=` Slope of the given line
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