KMadison
KMadison
06-02-2017
Mathematics
contestada
What is the answer to iii) ??
Respuesta :
arthurpdc
arthurpdc
06-02-2017
Since there is an expression in the square root, it must be non-negative. So:
[tex]f(x)=\sqrt{12-3x}\\\\
\Longrightarrow 12-3x\geq0\\\\
3x\leq12\\\\
x\leq\dfrac{12}{3}\\\\
x\leq4[/tex]
Then, the interval is:
[tex]
\boxed{x\in~]-\infty,4]}[/tex]
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