Question

Find the point at which the tangent to the curve y = x³-12x +18 are parallel to X - axis .​

Answer 1

The given equation of curve is

y = x³ -12x + 18

dy/dx = 3x² -12 {on differentiating wrt X }

so , the slope of line parallel to X axis .

°•° dy/dx = 0

=> 3x² - 12 = 0

=> x²= 12/3

=> X = +-2

For , X = 2

y = 2³- 12*2 + 18 = 2

and for X = - 2 , y = (-2)³ - 12(-2) + 18 = 34

so , the points are (2,2) and (-2,34) .

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Answer 2

Answer:

refer to this attachment ✅✅