函数f(x)=ax+b/1-x²,x∈(-1,1)满足f(0)=0,f(1/2)=2/5,求a,b的值并判断f(x)的奇偶性
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![函数f(x)=ax+b/1-x²,x∈(-1,1)满足f(0)=0,f(1/2)=2/5,求a,b的值并判断f(x)的奇偶性](/uploads/image/z/3816633-57-3.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dax%2Bb%2F1-x%26%23178%3B%2Cx%E2%88%88%EF%BC%88-1%2C1%EF%BC%89%E6%BB%A1%E8%B6%B3f%280%29%3D0%2Cf%281%2F2%29%3D2%2F5%2C%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC%E5%B9%B6%E5%88%A4%E6%96%ADf%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7)
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函数f(x)=ax+b/1-x²,x∈(-1,1)满足f(0)=0,f(1/2)=2/5,求a,b的值并判断f(x)的奇偶性
函数f(x)=ax+b/1-x²,x∈(-1,1)满足f(0)=0,f(1/2)=2/5,求a,b的值并判断f(x)的奇偶性
函数f(x)=ax+b/1-x²,x∈(-1,1)满足f(0)=0,f(1/2)=2/5,求a,b的值并判断f(x)的奇偶性
答:
f(x)=(ax+b)/(1-x²)
f(0)=b/(1-0)=b=0
f(1/2)=(a/2+0)/(1-1/4)=2/5
a=3/5
所以:a=3/5,b=0
f(x)=(3x/5)/(1-x²)=3x/(5-5x²)
定义域满足:1-x²≠0,x≠-1并且x≠1
定义域关于原点对称
f(-x)=-3x/(5-5x²)=-f(x)
所以:f(x)是奇函数