# What are the points of inflection, if any, of #f(x)=x^4-x^2+5 #?

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To find the points of inflection of the function ( f(x) = x^4 - x^2 + 5 ), we first find its second derivative, ( f''(x) ). Then, we set ( f''(x) = 0 ) and solve for ( x ). Finally, we check the concavity of the function around these points.

First derivative: [ f'(x) = 4x^3 - 2x ]

Second derivative: [ f''(x) = 12x^2 - 2 ]

Setting ( f''(x) = 0 ), we solve for ( x ): [ 12x^2 - 2 = 0 ] [ 12x^2 = 2 ] [ x^2 = \frac{1}{6} ] [ x = \pm \sqrt{\frac{1}{6}} ]

Since the second derivative changes sign at these points, ( x = \pm \sqrt{\frac{1}{6}} ), they are points of inflection.

Therefore, the points of inflection for the function ( f(x) = x^4 - x^2 + 5 ) are ( \left( -\sqrt{\frac{1}{6}}, f\left(-\sqrt{\frac{1}{6}}\right) \right) ) and ( \left( \sqrt{\frac{1}{6}}, f\left(\sqrt{\frac{1}{6}}\right) \right) ).

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