若已知f(x)=a^(2-3x) (a>0且a≠1),g(x)=a^x(1)求函数f(x)图象恒过定点坐标(2)求证:g[ (b+c)/2 ] ≤[ g(b)+a(c) ]/2ysf819036978的答案我看懂了、谢谢再问一题:y=a^(x^2+1) (a>0且a≠1)的值域
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![若已知f(x)=a^(2-3x) (a>0且a≠1),g(x)=a^x(1)求函数f(x)图象恒过定点坐标(2)求证:g[ (b+c)/2 ] ≤[ g(b)+a(c) ]/2ysf819036978的答案我看懂了、谢谢再问一题:y=a^(x^2+1) (a>0且a≠1)的值域](/uploads/image/z/11706972-60-2.jpg?t=%E8%8B%A5%E5%B7%B2%E7%9F%A5f%28x%29%3Da%5E%282-3x%29+%28a%3E0%E4%B8%94a%E2%89%A01%29%2Cg%28x%29%3Da%5Ex%281%29%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9B%BE%E8%B1%A1%E6%81%92%E8%BF%87%E5%AE%9A%E7%82%B9%E5%9D%90%E6%A0%87%282%29%E6%B1%82%E8%AF%81%3Ag%5B+%28b%2Bc%29%2F2+%5D+%E2%89%A4%5B+g%28b%29%2Ba%28c%29+%5D%2F2ysf819036978%E7%9A%84%E7%AD%94%E6%A1%88%E6%88%91%E7%9C%8B%E6%87%82%E4%BA%86%E3%80%81%E8%B0%A2%E8%B0%A2%E5%86%8D%E9%97%AE%E4%B8%80%E9%A2%98%EF%BC%9Ay%3Da%5E%28x%5E2%2B1%29+%28a%3E0%E4%B8%94a%E2%89%A01%29%E7%9A%84%E5%80%BC%E5%9F%9F)
若已知f(x)=a^(2-3x) (a>0且a≠1),g(x)=a^x(1)求函数f(x)图象恒过定点坐标(2)求证:g[ (b+c)/2 ] ≤[ g(b)+a(c) ]/2ysf819036978的答案我看懂了、谢谢再问一题:y=a^(x^2+1) (a>0且a≠1)的值域
若已知f(x)=a^(2-3x) (a>0且a≠1),g(x)=a^x
(1)求函数f(x)图象恒过定点坐标
(2)求证:g[ (b+c)/2 ] ≤[ g(b)+a(c) ]/2
ysf819036978的答案我看懂了、谢谢
再问一题:y=a^(x^2+1) (a>0且a≠1)的值域
若已知f(x)=a^(2-3x) (a>0且a≠1),g(x)=a^x(1)求函数f(x)图象恒过定点坐标(2)求证:g[ (b+c)/2 ] ≤[ g(b)+a(c) ]/2ysf819036978的答案我看懂了、谢谢再问一题:y=a^(x^2+1) (a>0且a≠1)的值域
(1) 因为任何数的0次方等于1,所以当2-3x=0 即 x=2/3时f(2/3)=a^0=1
所以函数f(x)图象恒过的定点坐标是(2/3,1)
(2)因为 g[(b+c)/2]=a^(b+c)/2=√a^(b+c)=√a^b•√a^c
[g(b)+g(c)]/2=(a^b+a^c)/2
所以 g[(b+c)/2] -[g(b)+g(c)]/2=0.5[2√a^b•√a^c-(a^b+a^c)]
= -0.5(√a^b-√a^c)^2≤0
所以 g[ (b+c)/2 ] ≤[ g(b)+g(c) ]/2
因为x^2+1≥1,(根据指数函数的单调性需要讨论)
(1)当a>1时,值域为【a,+∞) ( 单增 a^(x^2+1)≥a^1 )
(2) 当0<a<1时,值域为(0,a】