Question

find points at which tangent to the curve y=x³-3x²-9x+7 is parallel to the x axis​

Answer 1

Answer:

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)