Question

помогите плиз; 3+4x^2-x^4 найти критические точки

Answer 1

$f(x)=3+4x^2-x^4;\\f'(x)=(3+4x^2-x^4)'=(3)'+(4x^2)'-(x^4)'=0+4(x^2)'-4x^{4-1}=4*2x^{2-1}-4x^3=8x-4x^3;\\f'(x)=0;\\8x-4x^3=0;\\4x(2-x^2)=0;\\x_1=0;\\2-x^2=0;\\x^2=2;\\x_2=sqrt{2};\\x_3=-sqrt{2}$