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解:(1)f'(x)=1/(x+a)+2x依题意有f'(-1)=0,即a=3/2故f(x)=ln(x+3/2)+x^2从而f'(x)=(2x^2+3x+1)/(x+3/2)=(2x+1)(x+1)/(x+3/2)f(x)定义域为(-3/2,+无穷).当-3/20;当-1-1/2时,f'(x)>0.故f(x)分别在区间(-3/2,-1)、(-1/2,+无穷)上单调递增;在区间[-1,-1/2]上单调递减.(2)f(x)的定义域为(-a,+无穷),f'(x)=(2x^2+2ax+1)/(x+a)其中2x^2+2ax+1=0的判别式:4a... |
