已知函数y=f(x)满足f(x^2-5)=lg(x^2/(x^2-10))求解析式定义域判断奇偶性以及单调性
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已知函数y=f(x)满足f(x^2-5)=lg(x^2/(x^2-10))求解析式定义域判断奇偶性以及单调性
已知函数y=f(x)满足f(x^2-5)=lg(x^2/(x^2-10))求解析式定义域判断奇偶性以及单调性
已知函数y=f(x)满足f(x^2-5)=lg(x^2/(x^2-10))求解析式定义域判断奇偶性以及单调性
f(x^2-5)=lg(x^2/(x^2-10))则有
X^2-10>0,
X>√10或X5.
即,t∈(5,+∞),则有
f(t)=lg[(t+5)/(t-5)],(t∈(5,+∞),或t∈(-∞,-5)
则,函数y=f(x)解析式是:
f(x)=lg[(x+5)/(x-5)],(x∈(5,+∞),
定义域是:x∈(5,+∞),或X∈(-∞,-5)
f(-x)=lg[(-x+5)/(-x-5)]
=lg[(5-x)/(5+x)]
=lg(x-5)-lg(5+x)
=-lg(x+5)+lg(x-5)
=-[lg(x+5)-lg(x-5)]
=-lg(x+5)/(x-5)
=-f(x).
所以,f(x)在定义域范围内是奇函数.
令,X2>X1,X2-X1>0,
f(x2)-f(x1)
=lg[(x2+5)/(x2-5)]-lg[(x1+5)/(x1-5)]
=lg[(x2+5)(x1-5)/(x2-5)(x1+5)]
=lg[(x2+5)(x1-5)]-lg[(x2-5)(x1+5)].
又因为:[(x2+5)(x1-5)]-[(x2-5)(x1+5)]
=(X2-X1)(-5-5)
令x^2-5=t
f(x^2-5)=lg(x^2/(x^2-10))
f(t)=lg[(t+5)/(t-5)]=lg(t+5)-lg(t-5)
f(-t)=lg[(-t+5)/(-t-5)]
=lg[(t-5)/(t+5)]=lg(t-5)-lg(t+5)=-f(t)
为奇函数
设x1>x2
f(x1)-f(x2)
=lg(x1+5)...
全部展开
令x^2-5=t
f(x^2-5)=lg(x^2/(x^2-10))
f(t)=lg[(t+5)/(t-5)]=lg(t+5)-lg(t-5)
f(-t)=lg[(-t+5)/(-t-5)]
=lg[(t-5)/(t+5)]=lg(t-5)-lg(t+5)=-f(t)
为奇函数
设x1>x2
f(x1)-f(x2)
=lg(x1+5)-lg(x1-5)-lg(x2+5)+lg(x2-5)
=lg[(x1+5)*(x2-5)]-lg[((x2-5))*(x2+5)]
由于[(x1+5)*(x2-5)]>[((x2-5))*(x2+5)]
所以lg[(x1+5)*(x2-5)]>lg[((x2-5))*(x2+5)]
所以f(x1)>f(x2)
为增函数
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