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Solve for [tex]\( x \)[/tex]:
[tex]\[ 3x = 6x - 2 \][/tex]



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Problem Value: 1 point(s). Problem Score: [tex]$0 \%$[/tex]. Attempts Remaining: 8 attempts.

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Let [tex]$f(x)=\sqrt{2 x-3}$[/tex]
Use interval notation to indicate where [tex]$f(x)$[/tex] is continuous.
ANSWER: [tex]$\square$[/tex]
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Response:
Let [tex]\( f(x) = \sqrt{2x - 3} \)[/tex].

Use interval notation to indicate where [tex]\( f(x) \)[/tex] is continuous.

Answer: [tex]\( \square \)[/tex]



Answer :

GinnyAnswer
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]

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