If the definition field of function f (X & # 178; - 1) is [- 1,2], then the value range of X in function f (√ x) is X ∈ [- 1,2] → X & # 178; - 1 ∈ [- 1,3], ■ √ x ∈ [- 1,3] → x ∈ [0,9]. I don't understand how X & # 178; - 1 ∈ [- 1,3], and then get x ∈ [0,9] It's a bit confusing. Is the analytic expression of F (X & # 178; - 1) and f (√ x) the same, with the same range, but the whole (X & # 178; - 1, √ x) as X is different, so the definition field of X in that whole is required

If the definition field of function f (X & # 178; - 1) is [- 1,2], then the value range of X in function f (√ x) is X ∈ [- 1,2] → X & # 178; - 1 ∈ [- 1,3], ■ √ x ∈ [- 1,3] → x ∈ [0,9]. I don't understand how X & # 178; - 1 ∈ [- 1,3], and then get x ∈ [0,9] It's a bit confusing. Is the analytic expression of F (X & # 178; - 1) and f (√ x) the same, with the same range, but the whole (X & # 178; - 1, √ x) as X is different, so the definition field of X in that whole is required

X ∈ [- 1,2] is known
→ X & # 178; ∈ [0,4] square
→ X & # 178; - 1 ∈ [- 1,3] minus 1
That is, the value range of the whole in brackets is [- 1,3]
The overall range of √ x ∈ [- 1,3] is [- 1,3]
That is: √ x ∈ [0,3], √ x > = 0
The square of two sides of X ∈ [0,9]

If the odd function f (x) is an increasing function in the domain (- 1,1) and f (1 + a) + F (1-A & # 178;) < 0, the value range of a is obtained

Odd function, f (0) = 0;
It is also an increasing function in the domain of definition (- 1,1)
So f (x) is positive in (0,1) and negative in (- 1,0)
And f (x) + F (- x) = 0
So f (x) + F (y)

If the domain of function f (X & # 178; - 1) is [0,1], find the domain of function f (x)

If x belongs to (0,1), then (x ^ 2-1) belongs to (- 1,0), so the definition field is (- 1,0)