If it is known that f x x 2 1 and x belong to 1,1 , then the range of the function y f 2 x 1 is

Assuming that g(x)=2 x-1, first find the value range of g(x), and get that g(x)>-1, and at the same time, in f(g(x))=g(x) 2+1, the value range of g(x) is (-1,1), and g(x) is an increasing function with respect to x, then for f(g(x)), the value range of x is (-infinity, 1), and the value range of g(x) is (-1,1).

First, it is essential to determine the range of the value of x and the range of g(x), and then see that f(x) is a subtraction function on the interval of x (-1,0), and the value range is [1,2); is an increment function on [0,1) and the range is [1,2); then the range of f(x) on the interval (-1,1) with respect to x is [1,2);

Then determine the range of f(x).

In y=f(g(x))=g(x) 2+1, the range of x is (-infinity, 1), the range of the corresponding g(x) is (-1,1), and the range of f(g(x)) is [1,2].

It's complete

First find the range of 2 x-1, and we know that the range 2 x-1>-1, and in f(x)=x 2+1, x belongs to (-1,1), then 2 x-1 also belongs to (-1,1), and the intersection of the two domains is (-1,1), so y=f(2 x-1) is [1,2).

Substitute the value of x into the function, f(x) = 3,-1,1,3,5

So the range of f(x) is {1,3,5}

f(-1)= 2-1|=3

f(0)= 0-1|=1

f(1)=|2-1|=1

f(2)=|4-1|=3

f(3)=|6-1|=5

The value of f(x) is coarse =

Because the function f(x)=2x+1

So the function f(x 2 +1) = 2(x 2 +1) + 1 = 2x 2 +3 because x 2 0

So f(x 2 +1)=2x 2 +3 3 so the range of the function f(x 2 +1) is [3,+, so the answer is: [3,+

The domain of f(x) = (1-x 2) can be calculated as 1-x 2 >=0

-1 < = x< = 1 -1

(1-x^2)<=1

Root number 2< = x< = root number 2

So the domain of the function y=f(1-x 2) is [-root2, root2].

This is the hook function, from 0 to positive infinity there is a minimum value of 2, which is obtained when x=1, and f(1 2)=, so the value range is "2, thank you."

f'(x)=1-1 x 2=(x 2-1) x 2, so, when 01, the function increases, obtained by f(1 2)=1 2+2=5 2, f(1)=2, f(5)=5+1 5=26 5.

min=2,max=26 5, so the range of the function on [1 2,5] is: [2,26 5].