# What are the local extrema of #f(x)= 1/x-1/x^3+x^5-x#?

There are no local extrema.

graph{(5x^8-x^4-x^2+3)/x^4 [-5, 5, -10.93, 55]}

graph{x^-1-x^-3+x^5-x [-5, 5, -118.6, 152.4]}

No extrema!

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To find the local extrema of ( f(x) = \frac{1}{x} - \frac{1}{x^3} + x^5 - x ), we need to find where its derivative is zero.

[ f'(x) = -\frac{1}{x^2} + \frac{3}{x^4} + 5x^4 - 1 ]

Setting ( f'(x) = 0 ) and solving for ( x ), we get:

[ -\frac{1}{x^2} + \frac{3}{x^4} + 5x^4 - 1 = 0 ]

[ \frac{3}{x^4} - \frac{1}{x^2} + 5x^4 = 1 ]

[ 3 - x^2 + 5x^6 = x^2(5x^4 - 1) = 0 ]

This equation gives us the critical points where local extrema might occur. Solving ( 5x^4 - 1 = 0 ) gives ( x = \pm \sqrt[4]{\frac{1}{5}} ).

Testing the intervals defined by these critical points and the boundaries of the domain of ( f(x) ) to determine where ( f(x) ) is increasing or decreasing, and subsequently finding the local extrema:

- ( (-\infty, -\sqrt[4]{\frac{1}{5}}) ): ( f'(x) > 0 ), ( f(x) ) is increasing.
- ( (-\sqrt[4]{\frac{1}{5}}, \sqrt[4]{\frac{1}{5}}) ): ( f'(x) < 0 ), ( f(x) ) is decreasing.
- ( (\sqrt[4]{\frac{1}{5}}, +\infty) ): ( f'(x) > 0 ), ( f(x) ) is increasing.

Thus, ( f(x) ) has a local minimum at ( x = -\sqrt[4]{\frac{1}{5}} ) and a local maximum at ( x = \sqrt[4]{\frac{1}{5}} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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