Question

Help me to solve this question。

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

f(x)=x2−3bx+(c+2)f(1)=(1)2−3b(1)+

(c+2)=00=3−3b+c3b=(3+c)f′

(x)=2x−3bf′(3)=2(3)−3b=0b=23b−3=c

c=3

f(x)=x^2+3bx+c+2

f(1)=1^2+3b*1+c+2=0

1+3b+c+2=0

3b+c=-3…….eq(1)

f’(x)=2x+3b

f’(3)=2*3+3b=0

3b=-6

b=-2

Put this value of b in eq(1)

C=3

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