Key Concepts and Formulas

• Domain of $\cos^{-1}(u)$: The expression $\cos^{-1}(u)$ is defined for $-1 \le u \le 1$.
• Domain of $\log_a(u)$: The expression $\log_a(u)$ is defined for $u > 0$.
• Domain of $\frac{1}{g(x)}$: The expression $\frac{1}{g(x)}$ is defined for $g(x) \ne 0$.
• Absolute Value Inequalities: $|x| \le k$ implies $-k \le x \le k$.

Step-by-Step Solution

The given function is $f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _e(3-x)\right\}^{-1}$. To find the domain of $f(x)$, we need to consider the restrictions imposed by each part of the function.

Step 1: Domain restriction from the $\cos^{-1}$ term The term $\cos ^{-1}\left(\frac{2-|x|}{4}\right)$ is defined when its argument is between -1 and 1, inclusive. $-1 \le \frac{2-|x|}{4} \le 1$ Multiplying all parts by 4: $-4 \le 2-|x| \le 4$ Subtracting 2 from all parts: $-4 - 2 \le -|x| \le 4 - 2$ $-6 \le -|x| \le 2$ Multiplying all parts by -1 and reversing the inequality signs: $6 \ge |x| \ge -2$ This can be written as: $-2 \le |x| \le 6$ Since $|x| \ge 0$ for all real $x$, the condition $|x| \ge -2$ is always satisfied. Thus, the only effective restriction from this term is: $|x| \le 6$ This inequality is equivalent to $-6 \le x \le 6$. So, $x \in [-6, 6]$.

Step 2: Domain restrictions from the logarithmic term The term $\left\{\log _e(3-x)\right\}^{-1}$ can be written as $\frac{1}{\log_e(3-x)}$. This term imposes two restrictions:

Step 2a: Argument of the logarithm must be positive. The argument of the logarithm is $(3-x)$. Therefore, we must have: $3-x > 0$ $x < 3$ In interval notation, this is $x \in (-\infty, 3)$.

Step 2b: The denominator cannot be zero. The denominator is $\log_e(3-x)$. So, we must have: $\log_e(3-x) \ne 0$ The logarithm is zero when its argument is 1 ($e^0 = 1$). $3-x \ne 1$ $x \ne 3 - 1$ $x \ne 2$

Step 3: Combine all domain restrictions To find the domain of $f(x)$, we need to find the intersection of all the conditions derived:

1. $x \in [-6, 6]$ (from Step 1)
2. $x < 3$ (from Step 2a)
3. $x \ne 2$ (from Step 2b)

First, let's find the intersection of $x \in [-6, 6]$ and $x < 3$. This gives us $x \in [-6, 3)$. Now, we must also incorporate the restriction $x \ne 2$. Since $2$ is within the interval $[-6, 3)$, we must exclude it. Therefore, the domain of $f(x)$ is $[-6, 3) - \{2\}$.

Step 4: Identify $\alpha, \beta, \gamma$ and calculate $\alpha+\beta+\gamma$ The problem states that the domain of the function is $[-\alpha, \beta)-\{\gamma\}$. Comparing this with our derived domain $[-6, 3)-\{2\}$, we can identify: $-\alpha = -6 \implies \alpha = 6$ $\beta = 3$ $\gamma = 2$

Finally, we calculate $\alpha+\beta+\gamma$: $\alpha+\beta+\gamma = 6 + 3 + 2 = 11$

Common Mistakes & Tips

• Absolute Value Inequalities: Remember to correctly handle inequalities involving absolute values, especially when multiplying or dividing by negative numbers.
• Logarithm and Inverse Trigonometric Function Domains: Always recall the specific domain requirements for $\log_a(u)$ (argument $u>0$) and $\cos^{-1}(u)$ (argument in $[-1, 1]$).
• Excluding Values from Intervals: When a restriction like $g(x) \ne 0$ leads to excluding a specific value, ensure this value is removed from the combined interval, especially if it falls within that interval.

Summary We determined the domain of the function by analyzing the restrictions imposed by each component. The $\cos^{-1}$ term required its argument to be in $[-1, 1]$, leading to $x \in [-6, 6]$. The logarithmic term required its argument to be positive ($x < 3$) and its value to be non-zero, implying $x \ne 2$. Combining these conditions, the domain was found to be $[-6, 3) - \{2\}$. By matching this to the given format $[-\alpha, \beta)-\{\gamma\}$, we found $\alpha=6$, $\beta=3$, and $\gamma=2$. The sum $\alpha+\beta+\gamma$ is $6+3+2=11$.

The final answer is $\boxed{\text{11}}$.