What are the critical points of f(x)=7x^4-6x^2+1 f(x)=7x46x2+1?

1 Answer
Jun 10, 2018

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

Explanation:

f(x)=yf(x)=y

y=7x^4-6x^2+1y=7x46x2+1

dy/dx=28x^3-12xdydx=28x312x

dy/dx=0dydx=0

28x^3-12x=028x312x=0

7x^3-3x=07x33x=0

x(7x^2-3)=0x(7x23)=0

7x^2-3=07x23=0

7x^2=37x2=3

x^2=3/7x2=37

x=+-sqrt(3/7)x=±37

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

So the critical points of this equation is:

y=7(0.655)^4-6(0.655)^2+1=-0.286y=7(0.655)46(0.655)2+1=0.286
y=7(-0.655)^4-6(-0.655)^2+1=4.863y=7(0.655)46(0.655)2+1=4.863
y=7(0)^4-6(0)^2+1=1y=7(0)46(0)2+1=1

Critical points:

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