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

Question

1596 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-10T17:54:49

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

#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)#