指数函数y=1/2^X-1的值域是多少
解法一:反函数y=(1/2)^[(x-1)/(x 1)]logy=(x-1)/(x 1)xlogy log=x-1x(1-logy)=1 logyx=(1 logy)/(1-logy)反函数y=(1 logx)/(1-logx)反函数的定义域(0,1/2)∪(1/2, ∞)反函数的定义域即原函数的值域y的值域(0,1/2)∪(1/2, ∞)解法二:极限y的定义域(-∞,-1)∪(-1, ∞)y=(1/2)^[(x-1)/(x 1)]=2^[-(x-1)/(x 1)=2^{[2-(x 1)]/(x 1)}=2^[2/(x 1)-1]=2^[2/(x 1)]/2判断单调性x↗,x 1↗,2/(...全部
解法一:反函数y=(1/2)^[(x-1)/(x 1)]logy=(x-1)/(x 1)xlogy log=x-1x(1-logy)=1 logyx=(1 logy)/(1-logy)反函数y=(1 logx)/(1-logx)反函数的定义域(0,1/2)∪(1/2, ∞)反函数的定义域即原函数的值域y的值域(0,1/2)∪(1/2, ∞)解法二:极限y的定义域(-∞,-1)∪(-1, ∞)y=(1/2)^[(x-1)/(x 1)]=2^[-(x-1)/(x 1)=2^{[2-(x 1)]/(x 1)}=2^[2/(x 1)-1]=2^[2/(x 1)]/2判断单调性x↗,x 1↗,2/(x 1)↘,2^[2/(x 1)]/2↘即x↗,y↘y=在(-∞,-1)上单调递减,在(-1, ∞)上单调递减,判断定义域四个端点上y值的极限可得y的值域当x→-∞时,(x-1)/(x 1)→1 ,y→1/2-当x→-1-时,(x-1)/(x 1)→ ∞,y→0 y在(-∞,-1)上的值域是(0,1/2)当x→-1 时,(x-1)/(x 1)→-∞,y→ ∞当x→ ∞时,(x-1)/(x 1)→1-,y→1/2 y在(-1, ∞)上的值域是(1/2, ∞)y的值域(0,1/2)∪(1/2, ∞)。
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