Question

求下列函数的极值1.f(x)=bx²+x+22.f(x)=x³-12x3.f(x)=6-12x+x&

Answer 1

1.f(x)=bx²+x+2f(x)=b(x²+x/b+1/(2b)²-1/(2b)²)+2f(x)=b(x+1/(2b))²-1/(4b)+2x=-1/(2b) f(x)有极小值=-1/(4b)+22.f(x)=x³-12xf’(x)=3x²-12令f’(x)=03x²-12=0x=2 x=-2f’...