已知函数f(x)=2^x-1/[2^(绝对值x)]1)若f(x)=2,求x的值;(2)若2^t*f(2t)+mf(t)≥0对于t∈【1,2】恒成立,求实数m的取值范围.
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![已知函数f(x)=2^x-1/[2^(绝对值x)]1)若f(x)=2,求x的值;(2)若2^t*f(2t)+mf(t)≥0对于t∈【1,2】恒成立,求实数m的取值范围.](/uploads/image/z/2569260-12-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2%5Ex-1%2F%5B2%5E%EF%BC%88%E7%BB%9D%E5%AF%B9%E5%80%BCx%EF%BC%89%5D1%EF%BC%89%E8%8B%A5f%28x%29%3D2%2C%E6%B1%82x%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A52%5Et%2Af%282t%29%2Bmf%28t%29%E2%89%A50%E5%AF%B9%E4%BA%8Et%E2%88%88%E3%80%901%2C2%E3%80%91%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
已知函数f(x)=2^x-1/[2^(绝对值x)]1)若f(x)=2,求x的值;(2)若2^t*f(2t)+mf(t)≥0对于t∈【1,2】恒成立,求实数m的取值范围.
已知函数f(x)=2^x-1/[2^(绝对值x)]
1)若f(x)=2,求x的值;
(2)若2^t*f(2t)+mf(t)≥0对于t∈【1,2】恒成立,求实数m的取值范围.
已知函数f(x)=2^x-1/[2^(绝对值x)]1)若f(x)=2,求x的值;(2)若2^t*f(2t)+mf(t)≥0对于t∈【1,2】恒成立,求实数m的取值范围.
若x<=0,那个f(x)=0
f(x)=2,则x>0,
2^x-1/2^x=2
(2^x)^2-2*2^x-1=0
2^x=1+√2或2^x=1-√2
又x>0使得2^x>1,所以2^x=1+√2
x=log(2)(1+√2)
2^t(2^(2t)-1/2^(2t))+m(2^t-1/2^t)≥0
(2^t)^4+m*(2^t)^2-m-1≥0
(2^2t)^2+m*(2^2t)-m-1≥0
t∈[1,2],2^2t∈[4,16]
g(x)=x^2+mx-m-1=0
(x+m+1)(x-1)=0
则-m-1<4,m>-5
若x<=0,那个f(x)=0
f(x)=2,则x>0,
2^x-1/2^x=2
(2^x)^2-2*2^x-1=0
2^x=1+√2或2^x=1-√2
又x>0使得2^x>1,所以2^x=1+√2
x=log(2)(1+√2)
2^t(2^(2t)-1/2^(2t))+m(2^t-1/2^t)≥0
(2^t)^4+m*(2^t)^2-...
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若x<=0,那个f(x)=0
f(x)=2,则x>0,
2^x-1/2^x=2
(2^x)^2-2*2^x-1=0
2^x=1+√2或2^x=1-√2
又x>0使得2^x>1,所以2^x=1+√2
x=log(2)(1+√2)
2^t(2^(2t)-1/2^(2t))+m(2^t-1/2^t)≥0
(2^t)^4+m*(2^t)^2-m-1≥0
(2^2t)^2+m*(2^2t)-m-1≥0
t∈[1,2],2^2t∈[4,16]
g(x)=x^2+mx-m-1=0
(x+m+1)(x-1)=0
则-m-1<4,m>-5
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