函数单调性习题解答.1.若y=(2k+1)x+b是R上的减函数,则有( )2.已知函数f (x)在R上是增函数,若a + b>0,则( )A.f (a) + f (b)>f (-a) + f(-b) B.f (a) + f(b)>f (-a) – f(-b) C.f (a) + f (-a)>f (b) + f (-b) D
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 07:17:27
![函数单调性习题解答.1.若y=(2k+1)x+b是R上的减函数,则有( )2.已知函数f (x)在R上是增函数,若a + b>0,则( )A.f (a) + f (b)>f (-a) + f(-b) B.f (a) + f(b)>f (-a) – f(-b) C.f (a) + f (-a)>f (b) + f (-b) D](/uploads/image/z/6600322-10-2.jpg?t=%E5%87%BD%E6%95%B0%E5%8D%95%E8%B0%83%E6%80%A7%E4%B9%A0%E9%A2%98%E8%A7%A3%E7%AD%94.1.%E8%8B%A5y%3D%282k%2B1%29x%2Bb%E6%98%AFR%E4%B8%8A%E7%9A%84%E5%87%8F%E5%87%BD%E6%95%B0%2C%E5%88%99%E6%9C%89%EF%BC%88+%EF%BC%892%EF%BC%8E%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f+%28x%29%E5%9C%A8R%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E8%8B%A5a+%2B+b%EF%BC%9E0%2C%E5%88%99%EF%BC%88+%EF%BC%89A%EF%BC%8Ef+%28a%29+%2B+f+%28b%29%EF%BC%9Ef+%28-a%29+%2B+f%28-b%29+B%EF%BC%8Ef+%28a%29+%2B+f%28b%29%EF%BC%9Ef+%28-a%29+%E2%80%93+f%28-b%29+C%EF%BC%8Ef+%28a%29+%2B+f+%28-a%29%EF%BC%9Ef+%28b%29+%2B+f+%28-b%29+D)
xTN@~=ڲ7@9H}>@P*^ E4*HC @@JQ!`04kS^3ص*R_;7?;3 p5R38zeLn)PfygEXh}6[#fwYBm(\hbwkh%kn ~
zPPK%⪺: