Question

Use lmv to determine the point on thr curve y=x^3-3x where the tangent to the curve is parallel to the chord joining (1,-2),(2,2)

Answer 1

given equation of curve, y = x³ - 3x

differentiating y with respect to x,

dy/dx = 3x² - 3 = slope of tangent of curve {y = x³ - 3x}

a/c question,

tangent of curve is parallel to the chord joining (1, -2), (2, 2).

so, slope of chord = slope of tangent

so, slope of chord = {2 - (-2)}/{2 - 1} = 4/1

now, dy/dx = 3x² - 3 = 4

or, x² = 7/3

x = ±√{7/3}

at x = +√{7/3}

and y = x³ - 3x = 7/3√{7/3} - 3√{7/3} = -2/3√{7/3}

and at x = -{√7/3}

y = -7/3√{7/3} + 3√{7/3} = 2/3√{7/3}

hence, points are {√(7/3), -(2/3)√(7/3)}, {-√(7/3), (2/3)√(7/3)}