if f(x)+(x-3)^4(2x-1)^3, find the value(s) of x which solve f'(x)=0

Answer 1

f = (x-3)^4 * (2x+1)^3
f' = 4*(x-3)^3 * (2x+1)^3 + (x-3)^4 * 3 * (2x+1)^2 * 2
= 4(x-3)^3(2x+1)^3 + 6(x-3)^4(2x+1)^2
= 2(x-3)^3(2x+1)^2 (2(2x+1) + 3(x-3))
= 2(x-3)^3(2x+1)^2(4x+2+3x-9)
= 2(x-3)^3(2x+1)^2(7x-7)
= 14(x-3)^3(2x+1)^2(x-1)

So f' = 0 when x = -1/2 or 1 or 3