Calculus

Find the values of x that give relative extrema for the function f(x)=3x^5-5x^3
A. Relative maximum: x= 1; Relative minimum: x=sqrt(5/3)
B. Relative maximum: x=-1; Relative minimum: x=1
C. Relative maxima: x=+or- 1; Relative minimum: x=0
D. Relative maximum: x=0; Relative maxima: x=+or- 1
E. none of these

Solve without a graph

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  1. f ' (x) = 15x^4 - 15x^2
    = 0 for a max/min
    15x^2(x^2 - 1) = 0
    x=0 or x=±1

    to determine if these yield a max or a min look at the 2nd derivative
    f '' (x) = 60x^3 - 30x
    f "(0) = 0 , so x = 0 yields a point of inflection
    f "(1) = 60-30 > 0 , so x = 1 yields a minimum
    f "(-1) = -60 -(-30) < 0 , so x = -1 yields a maximum

    I think B fits my conclusions

    confirmation:
    http://www.wolframalpha.com/input/?i=plot+y+%3D3x%5E5-5x%5E3

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