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NTA AbhyasNTA Abhyas 2020Application of Derivatives
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Solution:
Let, f(x)=x5−5x4+5x3−1 ⇒f′(x)=5x4−20x3+15x2=0 ∴x2(x−3)(x−1)=0 or x=3,1,0
Now, f′′(x)=20x3−60x2+30x
Put x=3,1,0 , we get f′′(3)>0,f′′(1)<0,f′′(0)=0
Hence f(x) is minimum at x=3, maximum at x=1 , neither maximum nor minimum at x=0.