Find the stationary points of y = (x-7)(x-3)^2.

This problem requires the use of the product rule.Let u= x-7 ; by differentiation du/dx = 1. Let v = (x-3)^2 ; by differentiation using the chain rule, dv/dx = 2(x-3)
Product Rule: dy/dx = u*(dv/dx) + v*(du/dx)so dy/dx = (x-7)(2(x-3)) + (x-3)^2 = (2(x-7))*(x-3) + (x-3)^2= (x-3) (2(x-7)+(x-3))= (x-3) (3x-17)To find stationary points; dy/dx = 0. So x = 3 or 17/3
Substituting the x values into the original equations lets us find the co-ords of the stationary points ; which are :(3,0) & (5.67, -9.48) to 3 sf