We need
#(x^n)'=nx^(n-1)#
#a^2-b^2=(a-b)(a+b)#
Let #f(x)=x^4+2x^2-3#
We can factorise
#f(x)=(x^2+3)(x^2-1)=(x^2+3)(x+1)(x-1)#
#f'(x)=4x^3+4x=4x(x^2+1)#
We find critical points when #f'(x)=0#
#4x(x^2+1)=0# for #x=0#
We can calculate the second derivative
#f''(x)=12x^2+4#
#f''(0)=4 >0#, this is a minimum
We can do a sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaaaaaa)##0##color(white)(aaaaaaaa)##1##color(white)(aaaa)##+oo#
#color(white)(aaaa)##f'(x)##color(white)(aaaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)####color(white)(aaaa)##↘##color(white)(aaaaa)##-3##color(white)(aaaaaaaa)##↗#
#lim_(x->+-oo)f(x)=+oo#
graph{(y-(x^4+2x^2-3))=0 [-7.9, 7.9, -3.95, 3.95]}