已知函数f(x)=mx+n/1+x²是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17(1)确定函数解析式.(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数.(3)若实数t满足f(3t)+f(t+1)<0.求t的取值范围
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![已知函数f(x)=mx+n/1+x²是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17(1)确定函数解析式.(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数.(3)若实数t满足f(3t)+f(t+1)<0.求t的取值范围](/uploads/image/z/10133534-38-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dmx%2Bn%2F1%2Bx%26%23178%3B%E6%98%AF%E5%AE%9A%E4%B9%89%E5%9C%A8%5B-1%2F2%2C1%2F2%5D%E4%B8%8A%E7%9A%84%E5%A5%87%E5%87%BD%E6%95%B0%2C%E4%B8%94f%28-1%2F4%29%3D8%2F17%281%29%E7%A1%AE%E5%AE%9A%E5%87%BD%E6%95%B0%E8%A7%A3%E6%9E%90%E5%BC%8F.%EF%BC%882%EF%BC%89%E7%94%A8%E5%AE%9A%E4%B9%89%E8%AF%81%E6%98%8E%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B-1%2F2%2C1%2F2%5D%E4%B8%8A%E6%98%AF%E5%87%8F%E5%87%BD%E6%95%B0.%EF%BC%883%EF%BC%89%E8%8B%A5%E5%AE%9E%E6%95%B0t%E6%BB%A1%E8%B6%B3f%283t%29%2Bf%EF%BC%88t%2B1%29%EF%BC%9C0.%E6%B1%82t%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数f(x)=mx+n/1+x²是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17(1)确定函数解析式.(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数.(3)若实数t满足f(3t)+f(t+1)<0.求t的取值范围
已知函数f(x)=mx+n/1+x²是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17(1)确定函数解析式.
(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数.(3)若实数t满足f(3t)+f(t+1)<0.求t的取值范围
已知函数f(x)=mx+n/1+x²是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17(1)确定函数解析式.(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数.(3)若实数t满足f(3t)+f(t+1)<0.求t的取值范围
已知函数f(x)=(mx+n)/(1+x²)是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17.
(1)确定函数解析式。
f(-1/2)=(-m/2+n)/(1+1/4)
f(1/2)=(m/2+n)/(1+1/4)
f(-1/2)=-f(1/2)
(-m/2+n)/(1+1/4)=-(m/2+n)/(1+1/4)
-m/2+n=-m/2-n
n=0
f(x)=mx/(1+x²)
f(-1/4)=(-m/4)/(1+1/16)=-4m/17
f(-1/4)=8/17
-4m/17=8
m=-2
f(x)=-2x/(1+x²)
(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数。
1/2>x1>x2>-1/2
f(x1)-f(x2)=-2x1/(1+x1²)+2x2/(1+x2²)
=2(x2/(1+x2²)-x1/(1+x1²)
=2[x2(1+x1²)-x1(1+x2²)]/[(1+x2²)(1+x1²)]
(1+x2²)(1+x1²)>0
x2(1+x1²)-x1(1+x2²)
=x2+x1²x2-x1-x2²x1
=(x2-x1)-x1x2(x2-x1)
=(x2-x1)(1-x1x2)
x2-x1