作业帮1.若函数y=f(x)的定义域是[-2,4],则函数g(x)=f(x)+f(-x)的定义域是(B)A.[-4,4] B.[-2,2] C.[-4,-2] D[2,4]2.已知f(x)=ax2,g(x)=2/x,且f(2)=g(2),M={x｜f(x)>g(x)},则M等价于(C)A.[-2,2] B.(-2,2) C.(-∞,0)U(2,+∞) D.(-∞,0]U(2,+∞) 3.
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![1.若函数y=f(x)的定义域是[-2,4],则函数g(x)=f(x)+f(-x)的定义域是(B)A.[-4,4] B.[-2,2] C.[-4,-2] D[2,4]2.已知f(x)=ax2,g(x)=2/x,且f(2)=g(2),M={x｜f(x)>g(x)},则M等价于(C)A.[-2,2] B.(-2,2) C.(-∞,0)U(2,+∞) D.(-∞,0]U(2,+∞) 3.](/uploads/image/z/11966233-49-3.jpg?t=1.%E8%8B%A5%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AF%5B-2%2C4%5D%2C%E5%88%99%E5%87%BD%E6%95%B0g%28x%29%3Df%28x%29%2Bf%28-x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AF%28B%29A.%5B-4%2C4%5D+B.%5B-2%2C2%5D+C.%5B-4%2C-2%5D+D%5B2%2C4%5D2.%E5%B7%B2%E7%9F%A5f%28x%29%3Dax2%2Cg%28x%29%3D2%2Fx%2C%E4%B8%94f%282%29%3Dg%282%29%2CM%3D%7Bx%EF%BD%9Cf%28x%29%3Eg%28x%29%7D%2C%E5%88%99M%E7%AD%89%E4%BB%B7%E4%BA%8E%28C%29A.%5B-2%2C2%5D+B.%28-2%2C2%29+C.%28-%E2%88%9E%2C0%29U%282%2C%2B%E2%88%9E%29+D.%28-%E2%88%9E%2C0%5DU%282%2C%2B%E2%88%9E%29+3.)
1.若函数y=f(x)的定义域是[-2,4],则函数g(x)=f(x)+f(-x)的定义域是(B)A.[-4,4] B.[-2,2] C.[-4,-2] D[2,4]2.已知f(x)=ax2,g(x)=2/x,且f(2)=g(2),M={x｜f(x)>g(x)},则M等价于(C)A.[-2,2] B.(-2,2) C.(-∞,0)U(2,+∞) D.(-∞,0]U(2,+∞) 3.
1.若函数y=f(x)的定义域是[-2,4],则函数g(x)=f(x)+f(-x)的定义域是(B)
A.[-4,4] B.[-2,2] C.[-4,-2] D[2,4]
2.已知f(x)=ax2,g(x)=2/x,且f(2)=g(2),M={x｜f(x)>g(x)},则M等价于(C)
A.[-2,2] B.(-2,2) C.(-∞,0)U(2,+∞) D.(-∞,0]U(2,+∞)
3.函数f(x)=根号下mx2+4mx+m+3 的定义域为R,则实数m的取值范围是(B)
A.(0,1] B.[0,1] C.(-∞,0)U(1,+∞) D.(-∞,0)U[1,+∞)
1楼:
为什么"所以m>0,△=16m^2-4m(m+3)
1.若函数y=f(x)的定义域是[-2,4],则函数g(x)=f(x)+f(-x)的定义域是(B)A.[-4,4] B.[-2,2] C.[-4,-2] D[2,4]2.已知f(x)=ax2,g(x)=2/x,且f(2)=g(2),M={x｜f(x)>g(x)},则M等价于(C)A.[-2,2] B.(-2,2) C.(-∞,0)U(2,+∞) D.(-∞,0]U(2,+∞) 3.
1.显然,要使函数g(x)有定义,定义域是[-2,4]与[-4,2]的交集
所以选B
2.f(x)=ax2,g(x)=2/x,且f(2)=g(2)
所以a=1/4
所以f(x)>g(x)推出1/4x^2>2/x
解这个不等式得(-∞,0)U(2,+∞)
选(C)
3.函数f(x)=根号下mx2+4mx+m+3 的定义域为R,
所以mx2+4mx+m+3 恒大于等于0
所以m>0,△=16m^2-4m(m+3)