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