[tex]f\left(x\right)\:=\:x^5-2x^3+1[/tex][tex]f^{\prime}(x)=5x^4-6x^2[/tex][tex]\mathrm{Suppose\:that\:}x=c\mathrm{\:is\:a\:critical\:point\:of\:}f\left(x\right)\mathrm{\:then,\:}[/tex][tex]\mathrm{If\:}f\:'\left(x\right)>0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:maximum.}[/tex][tex]\mathrm{If\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)>\:0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:minimum.}[/tex][tex]\mathrm{If\:}f\:'\left(x\right)\mathrm{\:is\:the\:same\:sign\:on\:both\:sides\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:neither\:a\:local\:maximum\:nor\:a\:local\:minimum.}[/tex][tex]f^{\prime}(x)=5x^4-6x^2=0\text{ }\rightarrow x\text{ }^2(5x^2\text{ - 6})=0[/tex]
then at x = 0 and x = (6/5)^(1/2) the function has critical points. Minding the sign of f', one obtains that there are only local maximum and local minimum, so the answer is D.
