Question

Find critical numbers of the function
f(x)=x^4-4x^2

Answer 1

By the way this is a Calculus question.

Answer 2

differentiate it and put that eqn equal to zero, the zeroes of that eqn is are the critical points

Answer 3

i don't know what the answer for
x^2-2=0 is

Answer 4

take the first derivative of f(x) and you get:

f'(x)=4x^3-8x

then set that equal to zero

4x^3-8x=0

4x(x^2-2)=0 <--- Factored

4x=0 (x^2-2)=0
x=0 x^2=2
x= sqrt 2