Question

1 multiple choice 1 point which function does not have the set of all real numbers as its domain? f(x) = cos(x) + 1 f(x) = |2x - 1| $f(x) = \frac{x - 1}{x - 3}$ f(x) = 5^x - 3 clear my selection

Explanation:

Step1: Analyze \( f(x)=\cos(x)+1 \)

The cosine function \( \cos(x) \) is defined for all real numbers \( x \), so adding 1 doesn't change the domain. Domain: \( \mathbb{R} \).

Step2: Analyze \( f(x)=|2x - 1| \)

The absolute value function is defined for all real numbers \( x \) (since \( 2x-1 \) is linear, defined for all \( x \)). Domain: \( \mathbb{R} \).

Step3: Analyze \( f(x)=\frac{x - 1}{x - 3} \)

A rational function \( \frac{N(x)}{D(x)} \) is undefined when \( D(x)=0 \). Here, \( D(x)=x - 3 \), so \( x - 3 = 0\Rightarrow x = 3 \). Thus, \( x = 3 \) is excluded. Domain: \( \mathbb{R}\setminus\{3\} \), not all real numbers.

Step4: Analyze \( f(x)=5^{x}-3 \)

Exponential functions \( a^{x} \) ( \( a>0,a
eq1 \)) are defined for all real numbers \( x \), so \( 5^{x}-3 \) is defined for all \( x \). Domain: \( \mathbb{R} \).

Answer:

\( f(x)=\frac{x - 1}{x - 3} \) (the third option: \( f(x)=\frac{x - 1}{x - 3} \))