We have,
f(x)=x5−5x4+5x3−1
On differentiating and we get,
f′(x)=5x4−20x3+15x2
For maxima and minima
f′(x)=0
5x4−20x3+15x2=0
⇒5x2(x2−4x+3)=0
⇒x2−4x+3=0
So,
x2−(3+1)x+3=0
⇒x2−3x−1x+3=0
⇒x(x−3)−1(x−3)=0
⇒(x−3)(x−1)=0
If,
x−3=0
x=3
If,
x−1=0
x=1
Again differentiating and we get,
f′′(x)=20x3−60x2+30x
Put x=3
f′′(3)=20×(3)3−60×(3)2+30×3
f′′(3)=540−540+90
f′′(3)=90(minima)
If put x=1 and we get,
f′′(1)=20(1)3−60(1)2+30(1)
f′′(1)=−10(maxima)
Hence, this is the answer.