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The domain of the function [tex]f(x)=\log_{10}(6-|1-x|)[/tex] is the set of real values of [tex]x[/tex] for which:

A. [tex]-7 < x < 7[/tex]
B. [tex]-7 < x < 5[/tex]
C. [tex]-5 < x < 7[/tex]
D. [tex]-5 < x < 5[/tex]
E. None of the above.

Answer :

We begin by noting that the logarithm is only defined for positive arguments. Thus, for the function

[tex]$$
f(x)=\log_{10}(6-|1-x|)
$$[/tex]

the expression inside the logarithm must be positive:

[tex]$$
6-|1-x| > 0.
$$[/tex]

This inequality can be rearranged as:

[tex]$$
|1-x| < 6.
$$[/tex]

The inequality involving the absolute value can be written as a double inequality:

[tex]$$
-6 < 1-x < 6.
$$[/tex]

Now, we solve each part of the inequality separately.

1. For the left part:

[tex]$$
-6 < 1-x
$$[/tex]

Subtract 1 from both sides:

[tex]$$
-6-1 < -x \quad \Longrightarrow \quad -7 < -x.
$$[/tex]

Multiplying both sides by [tex]$-1$[/tex] (which reverses the inequality):

[tex]$$
7 > x \quad \Longrightarrow \quad x < 7.
$$[/tex]

2. For the right part:

[tex]$$
1-x < 6
$$[/tex]

Subtract 1 from both sides:

[tex]$$
-x < 5.
$$[/tex]

Again, multiply both sides by [tex]$-1$[/tex], remembering to reverse the inequality:

[tex]$$
x > -5.
$$[/tex]

Thus, we obtain the combined inequality:

[tex]$$
-5 < x < 7.
$$[/tex]

Therefore, the domain of the function is all real numbers [tex]$x$[/tex] such that:

[tex]$$
-5 < x < 7.
$$[/tex]

Among the provided options, the interval [tex]$-5

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