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If a series of functions have the same analytical formula, the same value range, but different definition fields, these functions are called "homologous functions", and the "homologous functions" with the analytical formula of y = x2 and the value range of {1,4} share () A. 7 B. 8 C. 9 D. 10
From the meaning of the question, it is known that functions of the same family are functions with different definition fields,
When the analytic formula of the function is y = x2 and the value range is {1, 4},
Its domain can be {1, 2} {1, - 2} {- 1, 2} {- 1, - 2} {1, - 1, 2} {1, - 1, - 2} {1, 2, - 2} {- 1, 2, - 2} {1, - 1, 2, - 2}
There are nine different situations,
Therefore, C
Find the value range of the following functions: (1) y = x ^ 2-3x + 1, X ∈ [0,3] (2) y = x + root sign (1-2x) (3) y = (3x + 2) / (x + 2)
【-0.25 1】
【0.5 1 】
[negative infinity positive infinity]
Find the value range of the following function (1) y = 2x + 3 / x-3 (2) y = x + 2x + 1 (3) y = 1 / 2x ^ 2-3x + 1 under the root sign
1) Y = 2x + 3 / x-3 = (2x-6 + 9) / x-3 = 2 + (9) / x-3 is the inverse proportional function type. It can be obtained that its value range is negative infinity to 2 and 2 to positive infinity. 2Y = x + squr (2x + 1) = ((squr (2X + 1) + 1) ^ 2) / 2-1, with squr 「 0, + infinity), then the value range is 「 - 1 / 2, + infinity). Squr refers to root 3) y = 1 / 2x ^ 2-3x + 1 = 1 / 2 (
Find the value range of function f (x) = root (- 2x + 3) - root (3x-4)
-2X + 3 decrement
So √ (- 2x + 3) decreases
And - √ (3x-4) decreases
So it's a subtraction function
-2x+3>=0,3x-4>=0
So 4 / 3
For the function y = (1 / 2) (x ^ 2-6x-7 in the root sign) power (1) find the definition field and value field of the function (2) find the monotone interval of the function Function y = (1 / 2) to the power of (x ^ 2-6x-7 in the root sign) (1) Find its definition field and value field? (2) Find its monotone interval?
Just let the part in the root sign ≥ 0. X ^ 2-6x-7 ≥ 0 (x + 1) (X-7) ≥ 0x ≤ - 1 or X ≥ 72. First, we divide the function into three parts. The value range of 1, G (x) = x ^ 2-6x-72, K (x) = G (x) 3 under the root sign, f (x) = (1 / 2) K (x) g (x) is [0, positive infinity]
Find the following function definition field, the X-2 power under the root sign of the value field y = (1 / 3)
The number in the root sign must be greater than or equal to 0, so X-2 is greater than or equal to 0 and X is greater than or equal to two
(I don't know if you're talking about 1 / 3 times or 1 / 3 power, so I'll calculate by 1 / 3 times!)
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