作业帮设函数f(x)=（x²/2）-ax+(a²-1/2),a∈R,若任取x∈[根号2,2],关于x的不等式f(x)≥(a²-4/2)恒成立求a的取值范围
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![设函数f(x)=（x²/2）-ax+(a²-1/2),a∈R,若任取x∈[根号2,2],关于x的不等式f(x)≥(a²-4/2)恒成立求a的取值范围](/uploads/image/z/13352994-18-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%EF%BC%88x%26%23178%3B%2F2%EF%BC%89-ax%2B%28a%26%23178%3B-1%2F2%29%2Ca%E2%88%88R%2C%E8%8B%A5%E4%BB%BB%E5%8F%96x%E2%88%88%5B%E6%A0%B9%E5%8F%B72%2C2%5D%2C%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x%29%E2%89%A5%28a%26%23178%3B-4%2F2%29%E6%81%92%E6%88%90%E7%AB%8B%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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设函数f(x)=（x²/2）-ax+(a²-1/2),a∈R,若任取x∈[根号2,2],关于x的不等式f(x)≥(a²-4/2)恒成立求a的取值范围
设函数f(x)=（x²/2）-ax+(a²-1/2),a∈R,若任取x∈[根号2,2],关于x的不等式f(x)≥(a²-4/2)恒成立求a的取值范围
设函数f(x)=（x²/2）-ax+(a²-1/2),a∈R,若任取x∈[根号2,2],关于x的不等式f(x)≥(a²-4/2)恒成立求a的取值范围
g(x)=f(x)-(a²-4/2)
=(x²/2)-ax+(a²-1/2)-(a²-4/2)
=x²/2-ax+3/2
=2(x²-2ax+a²)-a²/2+3/2
=2(x-a)²-a²/2+3/2
g(x)有最小值-a²/2+3/2
-a²/2+3/2≥0
-√3≤a≤√3