f(x)min=f(1)=1方法一：D(f)为x>=1,f'(x)=x在D内为单调递增函数,f'(x)min=f'(1)=1；f''(x)=根号下(x-1)在D内也为单调递增函数,f''(x)min=f''(1)=0,所以f(x)=f'(x)+f''(x)在定义域内的最小值是f(1)=1；方法二：D(f)为...