The local minimum of function is an argument x for which the first derivative of function g(x) is equal to zero, so: g'(x)=0 g'(x)=(x^4-5x^2+4)'=4x^3-10x=0 x(4x^2-10)=0 x=0 or 4x^2-10=0 4x^2-10=0 /4 x^2-10/4=0 x^2-5/2=0 [x-sqrt(5/2)][x+sqrt(5/2)]=0
Now we have to check wchich argument gives the minimum value from x=0, x=sqrt(5/2) and x=-sqrt(5/2). g(0)=4 g(sqrt(5/2))=25/4-5*5/2+4=4-25/4=-9/4 g(-sqrt(5/2))=-9/4 The answer is sqrt(5/2) and -sqrt(5/2).