Solution:
f(x)=3−5x3+5x4−x5
⇒f′(x)=−15x2+20x3−5x4
Put f′(x)=0
⇒−15x2+20x3−5x4=0⇒−5x2(3−4x+x2)=0
⇒−5x2=0,(3−4x+x2)=0
⇒x=0,(3−3x−x+x2)=0⇒(3−x)(1−x)=0⇒x=1,x=3,x=0
\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.}
We have, f′(x)>0 when 1<x<3 and f′(x)<0 when x<0,0<x<1,x>3, so x=3 is a point of local maximum and x=1 is a point of local minimum.
So, put x=3 in f(x).
f(3)=3−5(3)3+5(3)4−(3)5=30
And put x=1 in f(x).
f(1)=3−5(1)3+5(1)4−(1)5=2
Hence, local maximum is 30 at x=3 and minimum is 2 at x=1.
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