[tex] 1.\ t(x)=-\sqrt{x}\to y\leq0\\\\2.\ p(x)=\sqrt[3]{2-x}\to \text{All real numbers}\\\\3.\ w(x)=2+\sqrt{x}\to y\geq2\\\\4.\ r(x)=-2+\sqrt{2-x}\to y\geq-2\\\\5.\ k(x)=2-\sqrt{x}\to y\leq2\\\\6.\ v(x)=\sqrt{-x}\to y\geq0\\-------------------\\\sqrt{x}\geq0,\ \sqrt[3]{x}\in\mathbb{R} [/tex]
Answer:
[tex]1)\ t(x)=-\sqrt{x}------------\ \ y\leq 0\\\\2)\ p(x)=\sqrt[3]{2-x}-------\ \ \text{All real numbers}\\\\3)\ w(x)=2+\sqrt{x}---------\ \ y\geq 2\\\\4)\ r(x)=-2+\sqrt{2-x}--------\ \ y\geq -2\\\\5)\ k(x)=2-\sqrt{x}----------\ \ y\leq 2\\\\6)\ v(x)=\sqrt{-x}---------\ \ y\geq 0[/tex]
Step-by-step explanation:
1)
[tex]t(x)=-\sqrt{x}[/tex]
We know that:
[tex]\sqrt{x}\geq 0[/tex]
This means that:
[tex]-\sqrt{x}\leq 0[/tex]
( since when both side of the inequality is multiplied by a negative number then the sign of the inequality gets reversed )
This means that:
[tex]t(x)\leq 0[/tex]
Hence, the range is:
[tex]y\leq 0[/tex]
2)
[tex]p(x)=\sqrt[3]{2-x}[/tex]
We know that cube root is defined for all the real numbers and also the range covers the whole of R
Hence, the range is: All real numbers.
3)
[tex]w(x)=2+\sqrt{x}[/tex]
Again we know that:
[tex]\sqrt{x}\geq 0\\\\i.e.\\\\2+\sqrt{x}\geq 2\\\\i.e.\\\\w(x)\geq 2[/tex]
Hence, the range is:
[tex]y\geq 2[/tex]
4)
[tex]r(x)=-2+\sqrt{2-x}[/tex]
we know that:
[tex]\sqrt{x}\geq 0\\\\i.e.\\\\-2+\sqrt{x}\geq -2\\\\i.e.\\\\w(x)\geq -2[/tex]
Hence, the range is:
[tex]y\geq -2[/tex]
5)
[tex]k(x)=2-\sqrt{x}[/tex]
Since,
[tex]\sqrt{x}\geq 0[/tex]
This means that:
[tex]-\sqrt{x}\leq 0[/tex]
so,
[tex]-\sqrt{x}+2\leq 2\\\\i.e.\\\\2-\sqrt{x}\leq 2\\\\i.e.\\\\k(x)\leq 2[/tex]
Hence, the range is:
[tex]y\leq 2[/tex]
6)
[tex]v(x)=\sqrt{-x}[/tex]
We know that the square root of a number is always greater than or equal to zero.
so,
[tex]\sqrt{-x}\geq 0[/tex]
i.e.
The range is:
[tex]y\geq 0[/tex]
