What would the domain be for 2x-1/x+3 and x^2+9/4x^2-16
I think the answer is all real numbers but am not sure?
I think the answer is all real numbers but am not sure?
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2x - (1/x) + 3
The denominator can't be zero:
x != 0
Domain is: x can be anything except 0
Or did you mean:
(2x - 1) / (x + 3)
The denominator can't be zero:
x + 3 != 0
x != -3
Domain is: x can be anything except -3
x^2 + (9/4)x^2 - 16
Domain is all real numbers
Or did you mean:
x^2 + (9/(4x^2)) - 16
Or did you mean:
Domain is: x can be anything except zero
(x^2 + 9) / (4x^2 - 16)
The denominator can't be zero:
4x^2 - 16 != 0
4x^2 != 16
x^2 != 16/4
x^2 != 4
x != +/- sqrt(4)
x != +/- 2
Domain is: x can be anything except +/- 2
The denominator can't be zero:
x != 0
Domain is: x can be anything except 0
Or did you mean:
(2x - 1) / (x + 3)
The denominator can't be zero:
x + 3 != 0
x != -3
Domain is: x can be anything except -3
x^2 + (9/4)x^2 - 16
Domain is all real numbers
Or did you mean:
x^2 + (9/(4x^2)) - 16
Or did you mean:
Domain is: x can be anything except zero
(x^2 + 9) / (4x^2 - 16)
The denominator can't be zero:
4x^2 - 16 != 0
4x^2 != 16
x^2 != 16/4
x^2 != 4
x != +/- sqrt(4)
x != +/- 2
Domain is: x can be anything except +/- 2
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as you know domain is the input values for which the functions are defined the first function is not defined when x=0;as for x=0 1/x is not defined similarly for function 2 when x=+or- 2 the denominator becomes zero leading to a undefined value.. so the domain for 1 is R-{0} and for the second it is R-{-2,+2}...