Question

At what points will be tangents to the curve y = 2x³ − 15x² + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points.

Answer 1

Slope of x - axis is 0
Let (x₁, y₁) be the required point.
$$\begin{gathered} y=2x^3-15x^2+36x-21 \\ \mathrm{Since}\left(x_1,y_1\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}.\mathrm{Therefore} \\ {y}_1=2{x_1}^3-15{x_1}^2+36x_1-21...\left(1\right) \\ \mathrm{Now},y=2x^3-15x^2+36x-21 \\ \Rightarrow \frac{dy}{dx}=6x^2-30x+36 \\ \text{Slope of tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{= 6}{x_1}^2-30x_1+36 \\ \text{Given that} \\ \text{Slope of tangent at}\left(x,y\right)\text{= slope of the}\mathit{\text{x}}\text{-axis} \\ 6{x_1}^2-30x_1+36=0 \\ \Rightarrow {x_1}^2-5x_1+6=0 \\ \Rightarrow \left(x_1-2\right)\left(x_1-3\right)=0 \\ \Rightarrow x_1=2\text{or}{x}_1=3 \\ \text{Case-1:}{x}_1=2 \\ {y}_1=16-60+72-21=7\text{(From (1))} \\ \left(x_1,y_1\right)=\left(2,7\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-7=0\left(x-2\right) \\ \Rightarrow y=7 \\ \text{Case-2:}{x}_1=3 \\ {y}_1=54-135+108-21=6\text{(From (1))} \\ \left(x_1,y_1\right)=\left(3,6\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-6=0\left(x-3\right) \\ \Rightarrow y=6 \end{gathered}$$