To determine where the function [tex]\( f(x) = \sqrt{2x - 3} \)[/tex] is continuous, follow these steps:
1. Identify the domain of [tex]\( f(x) \)[/tex]:
The function [tex]\( f(x) \)[/tex] involves a square root, which means the expression inside the square root must be non-negative (since the square root of a negative number is not a real number). Therefore, we need to solve for [tex]\( x \)[/tex] in the inequality:
[tex]\[
2x - 3 \geq 0
\][/tex]
2. Solve the inequality:
Start by isolating [tex]\( x \)[/tex]:
[tex]\[
2x \geq 3
\][/tex]
Then divide both sides by 2:
[tex]\[
x \geq \frac{3}{2}
\][/tex]
3. Write the interval:
The solution [tex]\( x \geq \frac{3}{2} \)[/tex] means that [tex]\( x \)[/tex] must be at least [tex]\(\frac{3}{2}\)[/tex]. In interval notation, this is expressed as:
[tex]\[
\left[ \frac{3}{2}, \infty \right)
\][/tex]
Therefore, the function [tex]\( f(x) = \sqrt{2x - 3} \)[/tex] is continuous on the interval [tex]\(\left[ \frac{3}{2}, \infty \right)\)[/tex].
Final answer:
[tex]\[
\boxed{\left[ \frac{3}{2}, \infty \right)}
\][/tex]
