1.已知二次函数f(x),当x=1/2时,函数的最值为-49/4,且方程f(x)=0的两根之差为7,求f(x)的解析式.

1。f(x)=a(x-1/2)^2-49/4=ax^2-ax+(a-49)/4(x1-x2)^2=(x1+x2)^-4x1*x27^2=1^2-4*(a-49)/4aa=1f(x)=x^2-x-122。(1)2f(x)+3g(x)=9x^2-4x+12f(-x)+3g(-x)=9x^2+4x+1=-2f(x)+3g(x)两式相加6g(x)=18x^2+2g(x)=3x^2+1/3两式相减4f(x)=-8xf(x)=-2x(2)F(x)=[f(x)]^2-3g(x)=F(x)=[-2x]^2-3*(3x^2+1/3)=-5x^2-1值域是(负无穷,-1]单调递减区间是[0,正无穷)单调递增区间是（负无穷,0]3。函数f(x)= (ax^2+1) / (bx+c) 是奇函数f(-x)= (ax^2+1) / (-bx+c)=-f(x)= -(ax^2+1) / (bx+c)(ax^2+1) / (-bx+c)=-(ax^2+1) / (bx+c)(-bx+c)=-(bx+c)c=0f(x)= (ax^2+1) / (bx)≥2√ax/bx=2√a=2所以a=1f(x)= (x^2+1) /(bx)f(1)=2/b＜ 5/2b=1f(x)= (x^2+1) /x