Hi, Critical numbers will be where your derivative equals 0.

You are lucky enough that you already have f'(x), which is the derivative of some function f(x).

So all you need to do is find what x's make f'(x) = 0

So for example number 1,

solve 0 = 5x^4 -6x^2 + 1

this is easily done by making the substitution u = x^2

so

0 = 5u^2 - 6u +1

Solve by using the quadratic equation:

u = (6 +/- sqrt(36 - 20)) /10

u = (6+/- 4)/10

so u = 1 or 1/5

since x^2 = u

x^2 = 1 or 1/5

if x^2 = 1 then x = +/-1

if x^2 = 1/5 then x = +/-sqrt(1/5)

So you have a total of four critical points for number 1,

x= {-1, -sqrt(1/5), sqrt(1/5), 1}

try the others like this and see if you can get em.

good luck.