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

Finding the critical values requires that

Factor the left hand side.

Divide both sides by 4 to simplify.

Solve each term in the product separately.

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

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

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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.

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