Determine whether the function shown in the graph is even or odd.
2 answers:
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Here’s the answer hope it helps
A complement (
) is described by {x|x ∈ U and x is an even positive integer}
The function is 
.
The domain can only contain those values of x for which the denominator is not zero, that is all real numbers except 1,
AND for which 2-x i greater or equal to 0 (because if 2-x is negative, the square root of it cannot be calculated).
The last inequation,
, is solved as follows:
adding -2 to both sides we have
.
Multiplying by -1 (and not forgetting to swap the sign), we have
.
So x must be smaller or equal than 2, but also x cannot be 1, from the first condition. This means that the answer is <span>( - ∞ , 1 ) U ( 1 , 2].
</span>
Answer: ( - ∞ , 1 ) U ( 1 , 2]
The domain can only contain those values of x for which the denominator is not zero, that is all real numbers except 1,
AND for which 2-x i greater or equal to 0 (because if 2-x is negative, the square root of it cannot be calculated).
The last inequation,
adding -2 to both sides we have
Multiplying by -1 (and not forgetting to swap the sign), we have
So x must be smaller or equal than 2, but also x cannot be 1, from the first condition. This means that the answer is <span>( - ∞ , 1 ) U ( 1 , 2].
</span>
Answer: ( - ∞ , 1 ) U ( 1 , 2]