Question

Find points at which the tangent to the curve y = x 3 − 3x 2 − 9x + 7 is parallel to the x-axis.

Answer 1

$\text{\bf{when the tangent of curve is parallel}}\\\text{\bf{to the x- axis, then slope of tangent}}\\\text{\bf{equals to zero}}$

Here, y = x³ - 3x² - 9x + 7
differentiate y with respect to x
dy/dx = 3x² - 6x - 9
put dy/dx = 0 [ because slope of tangent , dy/dx is parallel to the x-axis ]
so, 3x² - 6x - 9 = 0
=> x² - 2x - 3 = 0
=> x² - 3x + x - 3 = 0
=> x(x - 3) + 1(x - 3) = 0
=> x = -1 , 3
at x = -1 , y = (-1)³ - 3.(-1)² - 9.(-1) + 7 = 12
at x = 3 , y = 3³ - 3.3² - 9.3 + 7 = -20

hence, required points are (-1,12) and (3,-20)

Answer 2

Answer:

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