find the minimum value of$$f(x)=x+\frac{1}{x}+x^2+\frac{1}{x^2}$$

Question

noelgreco · Answer

How can you find relative extreme values?

datanewb · Answer

Taking the derivative doesn't seem to help.
$$f'(x) = 1 - \frac{1}{x^2} + 2x - \frac{2}{x^3}$$ Only thing I can tell is that it's shooting towards infinity on either side of zero, and on the ends as well.  Interested in following this thread to see what the answer is. :) The minimum must be close to -1?

datanewb · Answer

f'(-1) = 0 and $$f''(x) = \frac{2}{x^3}+2+\frac{6}{x^4}\f''(-1) > 0$$, so -1 is at least a local minimum.

anonymous · Answer

$$t=x+\frac{1}{x}$$$|t|\ge2$$$f(t)=t^2+t-2$$finding minimum of this

datanewb · Answer

Well, the minimum of f(t) is -2, then
$$-2 = x + \frac{1}{x} \ x = -1,f(x) = 0$$ very  cool!
How much more difficult would it be to find the min of say $$f(x)=x+\frac{1}{x}+x^2+\frac{2}{x^2}$$?

anonymous · Answer

well this one is impossible to solve only with pen  and paper :)