已知函数f(x)=cx+1(0<c),2^(-x/2)(c≤x<1)满足f(c²)=9/8 1.求常数c的值2.解不等式f(x)>√2/8+1
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![已知函数f(x)=cx+1(0<c),2^(-x/2)(c≤x<1)满足f(c²)=9/8 1.求常数c的值2.解不等式f(x)>√2/8+1](/uploads/image/z/4493404-28-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dcx%2B1%EF%BC%880%EF%BC%9Cc%EF%BC%89%2C2%5E%EF%BC%88-x%2F2%EF%BC%89%EF%BC%88c%E2%89%A4x%EF%BC%9C1%EF%BC%89%E6%BB%A1%E8%B6%B3f%EF%BC%88c%26%23178%3B%EF%BC%89%3D9%2F8+1.%E6%B1%82%E5%B8%B8%E6%95%B0c%E7%9A%84%E5%80%BC2.%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%88x%EF%BC%89%EF%BC%9E%E2%88%9A2%2F8%2B1)
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已知函数f(x)=cx+1(0<c),2^(-x/2)(c≤x<1)满足f(c²)=9/8 1.求常数c的值2.解不等式f(x)>√2/8+1
已知函数f(x)=cx+1(0<c),2^(-x/2)(c≤x<1)满足f(c²)=9/8 1.求常数c的值
2.解不等式f(x)>√2/8+1
已知函数f(x)=cx+1(0<c),2^(-x/2)(c≤x<1)满足f(c²)=9/8 1.求常数c的值2.解不等式f(x)>√2/8+1
1、思路:0