Domain of arccos and logarithm reciprocal expression

Key concept: This question combines the $[-1,1]$ condition for arccos with the domain restrictions of a reciprocal logarithm.

Step by step

Step 1: Handle the arccos restriction

For

$f(x)=\cos^{-1}\left(\frac{2-|x|}{4}\right)+\{\log_e(3-x)\}^{-1},$

the argument of $\cos^{-1}$ must lie in $[-1,1]$:

$-1\le \frac{2-|x|}{4}\le 1.$

Multiply by 4:

$-4\le 2-|x|\le 4.$

The right inequality is always true because $2-|x|\le 2$.

The left inequality gives

$-4\le 2-|x|\quad\Rightarrow\quad |x|\le 6.$

So the arccos part requires

$-6\le x\le 6.$

Step 2: Handle the reciprocal logarithm restriction

The second term is

$\{\log_e(3-x)\}^{-1}=\frac{1}{\log_e(3-x)}.$

For this to be defined, we need:

1. $3-x>0$, so

$x<3;$

2. the denominator must not be zero, so

$\log_e(3-x)\ne 0.$

Since $\log_e(1)=0$, we must also exclude

$3-x\ne 1 \Rightarrow x\ne 2.$

Thus the second term requires

$x<3,\quad x\ne 2.$

Step 3: Intersect the conditions

Now intersect

$[-6,6]$ with

$(-\infty,3)\setminus\{2\}.$

This gives

$[-6,3)\setminus\{2\}.$

So the domain matches the form

$[-\alpha,\beta)-\{\gamma\}$

with

$\alpha=6,\quad \beta=3,\quad \gamma=2.$

Therefore,

$\alpha+\beta+\gamma=6+3+2=11.$

Why this works

Notice that the arccos term gives a closed interval, while the reciprocal logarithm removes one interior point and also forces an open upper bound at 3. The final domain is a closed interval with one missing point.

Pitfall alert

A very common mistake is to treat $\{\log_e(3-x)\}^{-1}$ as if only $3-x>0$ were needed. That is not enough: the logarithm must also be nonzero, or the reciprocal blows up. Another error is forgetting that $|x|\le 6$ comes from the arccos condition, not from the logarithm. Keep the two constraints separate first, then intersect them at the end.

Try different conditions

If the function were

$\cos^{-1}\left(\frac{2-|x|}{4}\right)+\{\log_e(4-x)\}^{-1},$

the arccos part would stay the same, but the logarithm restrictions would change to $x<4$ and $x\ne 3$. Intersecting with $[-6,6]$ would give $[-6,4)\setminus\{3\}$. The same method applies, but the removed point and endpoint move because the logarithm input changes.

Further reading

arccos domain, reciprocal logarithm, absolute value constraint