If the domain of F (X & # 178;) is [- 1,1], the domain of F (x) is
∵ f (X & # 178;) is defined as [- 1,1]
The definition field is the value range of X
The value range of X & # is [0,1]
And the range of values of ∵ () is the same
For f (x), X ∈ [0,1]
That is, the definition field is [0,1]
RELATED INFORMATIONS
- 1. Let f (x) and G (x) be the even and odd functions of R respectively in the domain of definition, f (x) = 2F (x) + G (x) = MX ^ 2 + NX + 1, f (1) = 1, f (- 1) = 5, M, n Finding f (2) and G (2)
- 2. Let even function f (x) satisfy f (x) = 2 & # 710; x-4 (x ≥ 0), then {x| f (X-2) > 0}= 1. {x | x 〈 - 2 or X 〉 4} 2. {x | x < 0 or X > 4} 3. {x | x < 0 or X > 6} 4. {x | x < - 2 or X > 2}
- 3. Let even function f (x) satisfy f (x) = x ^ 3-8 (x > = 0), if f (X-2) > 0, then the value range of X is? Did you answer that before? I can't see it. Can you answer it again
- 4. Which of the following functions are odd or even? Which are not odd or even? (1) f (x) = 5x + 3 (2) f (x) = 5x (3) f (x) = x & # 178; + 1
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- 11. 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
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