What are the values and types of the critical points, if any, of #f(x)=7x^4-6x^2+1 #?

1 Answer
Feb 3, 2017

#x=0# or #x=+-sqrt(3/7)#

Explanation:

Finding the critical values requires that

  • #f'(c)=0#, or
  • #f'(c)# is undefined, and
  • #c in # the domain of #f(x)#

Find the derivative of #f(x)#.

#f'(x)=28x^3-12x#

Set #f'(x)=0# and solve for #x#.

#28x^3-12x=0#

Factor the left hand side.

#4x(7x^2-3)=0#

Divide both sides by 4 to simplify.

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

Solve each term in the product separately.

#x=0" "# or #" "7x^2-3=0#
# " "7x^2=3#
#" "x^2=3/7#
#" "x=+-sqrt(3/7)#

Determine the types of critical points by looking at the graph of #f(x)#

graph{7x^4-6x^2+1 [-1.3, 1.3, -0.5, 1.5]}

When #x=0#, the critical point is a local maximum
When #x=+-sqrt(3"/"7)#, the critical points are local minimums

All three critical points stationary points because they have horizontal tangents at those critical points.