[ y = \arccos \left( \frac{2}{x} \right) + \sqrt{6 - x - x^{2}} ]
This is a combination of inverse trigonometric and square root functions. The domain of the function needs to be considered to avoid taking the arc cosine of an invalid argument and the square root of a negative number.
For the arc cosine function, the argument should be in the range ([-1, 1]), so ( \frac{2}{x} ) should be in that range. This implies that ( x \neq 0 ) and that (-1 \leq \frac{2}{x} \leq 1). From this, we can determine that ( -2 \leq x \leq 2 ) for the first part of the function to be defined.
For the square root to be real, the expression under the square root should be non-negative, so ( 6 - x - x^{2} \geq 0 ). Solving this inequality gives ( -3 \leq x \leq 2 ).
Therefore, the domain of the function is ([-3, 0) \cup (0, 2]).
The given function is:
[ y = \arccos \left( \frac{2}{x} \right) + \sqrt{6 - x - x^{2}} ]
This is a combination of inverse trigonometric and square root functions. The domain of the function needs to be considered to avoid taking the arc cosine of an invalid argument and the square root of a negative number.
For the arc cosine function, the argument should be in the range ([-1, 1]), so ( \frac{2}{x} ) should be in that range. This implies that ( x \neq 0 ) and that (-1 \leq \frac{2}{x} \leq 1). From this, we can determine that ( -2 \leq x \leq 2 ) for the first part of the function to be defined.
For the square root to be real, the expression under the square root should be non-negative, so ( 6 - x - x^{2} \geq 0 ). Solving this inequality gives ( -3 \leq x \leq 2 ).
Therefore, the domain of the function is ([-3, 0) \cup (0, 2]).