Question

question 6 solve please​

Answer 1

3 is the answer for 6 question

Answer 2

Answer:

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f(x)=x5−5x4+5x3−1

Differentiating w.r.t x we get,

f′(x)=5x4−20x3+15x2

f′(x)=5x2(x2−4x+3)

when f′(x)=0

⇒5x2(x2−4x+3)=0

⇒5x2(x−3)(x−1)=0

⇒x=0,x=3,x=1

Step 2

Difficult f′(x) again w.r.t x

f′′(x)=20x3−60x2+30x

when x=0;f′′(x)=0

Step 3

when x=3

f′′(x)=20(3)3−60(3)2+30(3)

=540−540+90

⇒f′′(x)>0

Hence x=3 is a point of local minimum.

Step 4

When x=1

f′′(x)=20(1)3−60(1)2+30(1)

=20−60+30

=−10

⇒f′′(x)<0

hence x=1 is the point of local maximum.

Step 5

Substituting x=3 in f(x) we get

f(3)=35−5(3)4+5(3)3−1

=243−405+135−1

=−28

f(1)=15−5(1)4+5(1)3−1

=1−5+5−1

=0

Hence local maximum =0

local minimum =−28

x=0 is the point of inflection.