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On the mathematical problems of derivatives ~ Solve ah ah ~
f′(x)= 3ax? +2x+b

g(x)= f(x)+f′(x)= ax? +x? +bx+3ax? +2x+b =ax? +(3a+ 1)x? +(2+b)x+b

G(x)= odd function.

g(-x)=-g(x)

therefore

b=0

3a+ 1=0

a=- 1/3

therefore

f(x)=- 1/3x? +x?

g(x)==ax? +(3a+ 1)x? +(2+b)x+b

=- 1/3x? +2 times

g( 1)=- 1/3+2=5/3

g(2)=-8/3+4=4/3

The derivative of g(x) =-x? +2=0

X= radical number 2= 1.4

X=- root number 2=- 1.4 (deleted)

g( 1.4)=- 1/3( 1.4)? +2( 1.4)= 1.9

(Calculated by root number 2 is 2/3 of root number 4)

So the maximum value is g( 1.4)= 1.9.

(Calculated by root number 2 is 2/3 of root number 4)

The minimum value is g(2)=4/3.