I have to minimize the function $ƒ(x_1,x_2)=(x_1-1)^2+x_2^3-x_1x_2$. The initial point is $[1,1]^T$.
The gradient of this function is $∇ƒ(x_1,x_2)=[2(x_1-1)-x_2,3x_2^2-x1]$. This gradient evaluated in the initial point is $∇ƒ(1,1)=[-1,2]$. Following the steepest descent method it is mandatory to minimize the function $ƒ(x_0-α∇ƒ(x_0))$ in order to find the value of $α$. So $ƒ(x_0-α∇ƒ(x_0))=-5α+15α^2-8α^3$ and $ƒ'(x_0-α∇ƒ(x_0))=-5+30α-24α^2$. This function has extreme points in $α_1=0.95061$ and $α_2=5.094$. In order to be a minimum of this curve $ƒ''(x_0-α∇ƒ(x_0))=30-48α$ has to be positive. This is my problem $ƒ''(x_0-α∇ƒ(x_0))$ evaluated at both $α$ values is negative so they don't minimize the direction. So what I am doing wrong?