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

Question

Emma Aldridge · Accepted Answer

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

Finding the critical values requires that

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.

Emilia Aguilar · Answer

undefined To find the critical points of $ f(x) = 7x^4 - 6x^2 + 1 $, we need to find where the derivative is equal to zero.

First, compute the derivative:
$$ f'(x) = 28x^3 - 12x $$

Now, set $ f'(x) $ equal to zero and solve for $ x $:
$$ 28x^3 - 12x = 0 $$
$$ 4x(7x^2 - 3) = 0 $$

This equation yields critical points at $ x = 0 $ and $ x = \pm \sqrt{\frac{3}{7}} $.

To determine the type of each critical point, we can use the second derivative test.

Second derivative:
$$ f''(x) = 84x^2 - 12 $$

Evaluate $ f''(0) $, $ f''(\sqrt{\frac{3}{7}}) $, and $ f''(-\sqrt{\frac{3}{7}}) $ to determine the concavity at each critical point.

$$ f''(0) = -12 $$ which implies a maximum.

$$ f''\left(\sqrt{\frac{3}{7}}\right) = 60 $$ which implies a minimum.

$$ f''\left(-\sqrt{\frac{3}{7}}\right) = 60 $$ which also implies a minimum.