Find and characterise the critical points of: $f(x)=(2x^3-12x^2+18x-1)^5$
I differentiated to get:
$$\frac{df}{dx}=5(6x^2-24x+18)(2x^3-12x^2+18x-1)^4=30(x-3)(x-1)(2x^3-12x^2+18x-1)^4$$
But don't know how to factorise this further?
Find and characterise the critical points of: $f(x)=(2x^3-12x^2+18x-1)^5$
I differentiated to get:
$$\frac{df}{dx}=5(6x^2-24x+18)(2x^3-12x^2+18x-1)^4=30(x-3)(x-1)(2x^3-12x^2+18x-1)^4$$
But don't know how to factorise this further?
If you set $$30(x-3)(x-1)(2x^3-12x^2+18x-1)^4=0$$ then the stationary points are at $$x-3=0\implies x=3\\x-1=0\implies x=1\\2x^3-12x^2+18x-1=0$$ and this cubic equation can be solved using Cardano's method.
To find the critical points, substitute each value of $x$ into $y=(2x^3-12x^2+18x-1)^5$ and the nature of them can be found by evaluating the second derivative of $y$ at those points.