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现行的参考书和许多数学刊物,都不时出现求值域的一种方法——根据反函数的定义域求原函数的值域,即:要求y=f(x)的值域,可先求出y=f~(-1)(x),y=f~(-1)(x)的定义域即为y=f(x)的值域。例求函数y=x+6(x-9)~(1/2)-1的值域。解 y=((x-9)~(1/2))~2+2·3·(x-9)~(1/2)+3~2-1=((x-9~(1/2))+3)~2-1 ∴ (y+1)~(1/2)=(x-9)~(1/2)+3, (x-9)~(1/2)=(y+1)~(1/2)-3,x=y-6(y+1)~(1/2)+19。所给函数的反函数为y=x-6(x+1)~(1/2)+19。其定义域[-1,+∞)即为所求值域。
The current reference books and many mathematics publications, from time to time there is a way to evaluate the domain - according to the domain of the inverse function to find the value of the original function, that is: y = f (x) of the required value range, can be the first The domain of y=f~(-1)(x),y=f~(-1)(x) is the domain of y=f(x). Example Find the range of the function y=x+6(x-9)~(1/2)-1. Solution y=((x-9)~(1/2))~2+2•3(x-9)~(1/2)+3~2-1=((x-9~(1/ 2)) +3)~2-1 ∴ (y+1)~(1/2)=(x-9)~(1/2)+3, (x-9)~(1/2)=( y+1)~(1/2)-3, x=y-6(y+1)~(1/2)+19. The inverse function of the given function is y=x-6(x+1)~(1/2)+19. Its definition domain [-1, +∞) is the value domain.