What are the critical points of f(x)=7x^4-6x^2+1 ?

1 Answer
Jun 10, 2018

(0.655, -0.286)
(-0.655, 4.863)
(0, 1)

Explanation:

f(x)=y

y=7x^4-6x^2+1

dy/dx=28x^3-12x

dy/dx=0

28x^3-12x=0

7x^3-3x=0

x(7x^2-3)=0

7x^2-3=0

7x^2=3

x^2=3/7

x=+-sqrt(3/7)

x=0.655 and x=-0.655 and x=0

So the critical points of this equation is:

y=7(0.655)^4-6(0.655)^2+1=-0.286
y=7(-0.655)^4-6(-0.655)^2+1=4.863
y=7(0)^4-6(0)^2+1=1

Critical points:

(0.655, -0.286)
(-0.655, 4.863)
(0, 1)