At what points will the tangent to the curve y=2x^3-15x^2+36x-21 be parallel to the x axis Also, find the - Maths - Application of Derivatives

Question

At what points will the tangent to the curve y=2x^3-15x^2+36x-21 be
parallel to the x axis? Also, find the equations of tangents to the
curve at these points

Mayur Pisode · Accepted Answer

Let the required point be Px1,y1
Now, dydxx1,y1 = 0Now, y = 2x3 - 15x2+36x - 21⇒dydx = 6x2-30x + 36 = 6x2-5x+6⇒dydxx1,y1 = 6x12-5x1+6Now,  dydxx1,y1= 0⇒6x12-5x1+6 = 0⇒x1-2x1-3 = 0⇒x1 = 2  and  x1 = 3Since,  Px1,y1 lies on the given curve, so we havey1 = 2x13 - 15x12+36x1 - 21When x1 = 2y1 = 223 - 1522 + 362 - 21 = 7When x1 = 3y1 = 233 - 1532 + 363 - 21 =6So, required points are 2,7 and 3,6The equation of the tangent at 2,7 is y = 7The equation of the tangent at 3,6 is y = 6

At what points will the tangent to the curve y=2x^3-15x^2+36x-21 be parallel to the x axis? Also, find the equations of tangents to the curve at these points.