What are the critical values, if any, of #f(x)=x^5-5x^3+2x#?

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Answer 1 · 2016-03-03T16:40:03

They are #+-sqrt((15+-6sqrt5)/10)#

The domain of #f# is all real numbers.

#f'(x) = 5x^4-15x^2+2#

#f'# never fails to exists, so the critical values are the zeros of #f'#

#5x^4 -15x^2+2# is quadratic in #x^2#, so the roots can be found by using the quadratic formula.

#x^2 = (-(-15)+-sqrt((-15)^2-4(5)(2)))/(2(5)) #

# = (15+-sqrt180)/10#

Since #sqrt180 < 15#, both values are positive, so we get

#x= +-sqrt((15+-6sqrt5)/10)#