How do you sketch the graph #y=x^4+2x^2-3# using the first and second derivatives?

1 Answer
Jan 1, 2017

See the sketch of the graph below

Explanation:

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]}