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Find points at which the tangent to the curve `y=x^3-3x^2-9x+7`is parallel to the x-axis. |
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Answer» Equation of curve ` y = x^(3) - 3x^(2) - 9x +7`…(1) `rArr (dy)/(dx)= 3x^(2) - 6x-9` The tangent of the curve will be parallel to x-axis at those point where slope = 0. `:. 3x^(2) - 6x - 9 =0` `rArr x^(2) - 2x - 3 = 0` ` rArr(x-3)(x+1) = 0` ` rArr x= 3 or x =- 1` put x = 3 in equation (1) ` y = 27 - 27 - 27 +7 =-20` Put x =- 1 in equation (1) ` y =- 1- 3+9+7=12` `:.` Required point = (3, -20) and ( -1, 12) |
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