第一问
f(x)=(3/8)x^2-2x+2+lnx
f'(x)=(3/4)x-2+1/x
f''(x)=3/4-1/x^2
即当f''(x)=0,即3/4-1/x^2=0,x=±2√3/3时,存在拐点,这时候取极值
当f''(x)>0,即3/4-1/x^2>0,即x>2√3/3或x
第一问
f(x)=(3/8)x^2-2x+2+lnx
f'(x)=(3/4)x-2+1/x
f''(x)=3/4-1/x^2
即当f''(x)=0,即3/4-1/x^2=0,x=±2√3/3时,存在拐点,这时候取极值
当f''(x)>0,即3/4-1/x^2>0,即x>2√3/3或x