Given: A real-valued function
[tex]f(x)=-3+\sqrt{4x-12}[/tex]Required: Domain and Range of the function.
Explanation:
Domain of the function is all the values of x, for which the function is defined.
Here, root function is defined when 4x-12 is greater than equal to zero.
So domain is
[tex]\begin{gathered} 4x-12\ge0 \\ 4x\ge12 \\ x\ge3 \end{gathered}[/tex]Thus, domain is
[tex]x\ge3[/tex]Now, for range
[tex]\begin{gathered} 3\leq x<\infty \\ 12\leq4x<\infty \end{gathered}[/tex]Further, subtracting 12
[tex]\begin{gathered} 0\leq4x-12<\infty \\ 0\leq\sqrt{4x-12}<\infty \end{gathered}[/tex]Adding -3
[tex]\begin{gathered} -3\leq-3+\sqrt{4x-12}<\infty \\ -3\leq f(x)<\infty \end{gathered}[/tex]Thus range is
[tex]f(x)\ge-3[/tex]Final answer: Option 3 is correct answer.
