\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}
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![\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}](/uploads/image/z/13415075-35-5.jpg?t=%5Cbegin%7Barray%7D%7Bl%7D%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx-%5Cfrac%7B1%7D%7Bx%2B1%7D%2Cg%28x%29%3D%7Bx%7D%5E%7B2%7D-2ax%2B4%2C%E8%8B%A5%26%238704%3B%7Bx%7D_%7B1%7D%E2%88%88%5B0%2C1%5D%2C%26%238707%3B%7Bx%7D_%7B2%7D%E2%88%88%5B1%2C2%5D%2C%5C%5C+%E4%BD%BFf%28%7Bx%7D_%7B1%7D%29%E2%89%A5g%28%7Bx%7D_%7B2%7D%29%2C%E5%88%99%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%5Cend%7Barray%7D)
\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}
\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}
\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}
f(x)=x-1/(1+x)在[0,1]单调增加 其最小值为f(0)=-1
故g(x)=x^2-2ax+4≤-1 在[1,2]恒成立 令x=1 可得到 a≥3>2 故g(x)在[1,2]单调减小 只需要g(1)≤-1
解得 a≥3