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
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
1 answer