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

Question

Scarlett Allison · Accepted Answer

There are no local extrema.

Local extrema could occur when #f'=0# and when #f'# switches from positive to negative or vice versa. #f(x)=x^-1-x^-3+x^5-x# #f'(x)=-x^-2-(-3x^-4)+5x^4-1# Multiplying by #x^4/x^4#: #f'(x)=(-x^2+3+5x^8-x^4)/x^4=(5x^8-x^4-x^2+3)/x^4# Local extrema could occur when #f'=0#. Since we can't solve for when this happens algebraically, let's graph #f'#: #f'(x)#: graph{(5x^8-x^4-x^2+3)/x^4 [-5, 5, -10.93, 55]} #f'# has no zeros. Thus, #f# has no extrema. We can check with a graph of #f#: graph{x^-1-x^-3+x^5-x [-5, 5, -118.6, 152.4]} No extrema!

Emma Aldridge · Answer

undefined 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}} $.