The domain of the function(D) is the set of all values that argument might take. I will assume that x is a real number. Also the conditions is inaccurate, so i don't fully sure whether i recognized it correctly in each case, but it must be close to it
f(x) = x + 10. D : x β R D:x\isin R D : x β R
A ( x ) = x 2 β 2 A(x) = x^2 -2 A ( x ) = x 2 β 2 . D : x β R D:x\isin R D : x β R
F ( x ) = 2 3 π₯ + 5 F(x) = 2^{3π₯} + 5 F ( x ) = 2 3 x + 5 . D : x β R D:x\isin R D : x β R
H ( x ) = π₯ β 2 H(x) = \sqrtπ₯ β 2 H ( x ) = x β β 2 . The value under the square root must be non-negative, so D : x β [ 0 , + β ) D:x\isin [0,+\infty) D : x β [ 0 , + β )
g(x) = 5 β 3x . D : x β R D:x\isin R D : x β R
K ( x ) = x 2 β 2 K(x) = \sqrt{x^2 β 2} K ( x ) = x 2 β 2 β . D : x 2 β 2 β₯ 0 β
β βΉ β
β x 2 β₯ 2 β
β βΉ β
β D : x β ( β β , β 2 ) βͺ ( 2 , + β ) D:x^2-2β₯0\implies x^2β₯2\implies D:x\isin (-\infty,-\sqrt2)\cup(\sqrt2,+\infty) D : x 2 β 2 β₯ 0 βΉ x 2 β₯ 2 βΉ D : x β ( β β , β 2 β ) βͺ ( 2 β , + β )
g ( x ) = 1 ( π₯ + 5 ) ( π₯ β 1 ) g(x) = {\frac 1 {(π₯+5)(π₯β1)}} g ( x ) = ( x + 5 ) ( x β 1 ) 1 β . Cannot divide by 0, so D : ( x + 5 ) ( x β 1 ) =ΜΈ 0 β
β βΉ β
β D : x β R D:(x+5)(x-1)\not=0\implies D:x\in R D : ( x + 5 ) ( x β 1 ) ξ = 0 βΉ D : x β R \ {-5, 1}
C ( x ) = 2 x 3 + 4 x 2 β 2 x + 1 C(x) = 2x^3 + 4x^2 - 2x + 1 C ( x ) = 2 x 3 + 4 x 2 β 2 x + 1 . D : x β R D:x\isin R D : x β R
b ( x ) = π₯ β 1 x 2 + 5 x + 6 b(x) = π₯β{\frac 1 {x^2+5x+6}} b ( x ) = x β x 2 + 5 x + 6 1 β . D : x 2 + 5 x + 6 =ΜΈ 0 β
β βΉ β
β D : x β R D:x^2+5x+6\not=0\implies D:x\in R D : x 2 + 5 x + 6 ξ = 0 βΉ D : x β R \ {-3, -2}
f ( x ) = x β 1 x β 2 f(x)={\frac {\sqrt{x-1}} {x-2}} f ( x ) = x β 2 x β 1 β β . D : ( x β 1 β₯ 0 ) β§ ( x β 2 =ΜΈ 0 ) β
β βΉ β
β D : x β [ 1 , + β ) D:(x-1β₯0)\land (x-2\not=0)\implies D:x\isin [1,+\infty) D : ( x β 1 β₯ 0 ) β§ ( x β 2 ξ = 0 ) βΉ D : x β [ 1 , + β ) \ {2}
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