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Given the function f (x) = 2lnx + x2 + ax, if the curve y = f (x) has a tangent parallel to the straight line 2x-y = 0, then the value range of real number a is () A. （-∞，-2]B. （-∞，-2）C. （-2，+∞）D. [-2，+∞）
The function f (x) = 2lnx + x2 + ax has a tangent parallel to the line 2x-y = 0, that is, f ′ (x) = 2 has a solution on (0, + ∞), while f ′ (x) = 2 · LX + 2x + A, that is, 2x + 2X + a = 2 has a solution on (0, + ∞), a = 2-2 (x + 1x). Because x > 0, when x + 1x ≥ 2, x = 1, the equal sign holds, that is, there is a ≤ 2-4, so the value range of a is (- ∞, - 2]. So a is selected
Is the domain of the derivative function y = 1 / X of y = LNX x x > 0?
Yes, but the exact writing should be (0, + ∞)
Because the domain of the original function is (0, + ∞), the derivative function is only differentiable at (0, + ∞)
What is the domain of the function y = (X-2) / LNX + + √ (9-x ^ 2)?
If x is the true number of LNX, then x > 0; (1)
If LNX is the denominator, then LNX is not 0, that is, X ≠ 1; (2)
If 9-x ^ 2 is squared, then 9-x ^ 2 > = 0, so, - 3
The definition field of function y = (1 / LNX + 1) - √ 3-x is ()
3-x>=0
x
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