Question

Match each function with its range.

Answer 1

Answer:

$1)\ t(x)=-\sqrt{x}------------\ \ y\leq 0\\\\2)\ p(x)=\sqrt[3]{2-x}-------\ \ \text{All real numbers}\\\\3)\ w(x)=2+\sqrt{x}---------\ \ y\geq 2\\\\4)\ r(x)=-2+\sqrt{2-x}--------\ \ y\geq -2\\\\5)\ k(x)=2-\sqrt{x}----------\ \ y\leq 2\\\\6)\ v(x)=\sqrt{-x}---------\ \ y\geq 0$

Step-by-step explanation:

1)

$t(x)=-\sqrt{x}$

We know that:

$\sqrt{x}\geq 0$

This means that:

$-\sqrt{x}\leq 0$

( since when both side of the inequality is multiplied by a negative number then the sign of the inequality gets reversed )

This means that:

$t(x)\leq 0$

Hence, the range is:

$y\leq 0$

2)

$p(x)=\sqrt[3]{2-x}$

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)

$w(x)=2+\sqrt{x}$

Again we know that:

$\sqrt{x}\geq 0\\\\i.e.\\\\2+\sqrt{x}\geq 2\\\\i.e.\\\\w(x)\geq 2$

Hence, the range is:

      $y\geq 2$

4)

$r(x)=-2+\sqrt{2-x}$

we know that:

$\sqrt{x}\geq 0\\\\i.e.\\\\-2+\sqrt{x}\geq -2\\\\i.e.\\\\w(x)\geq -2$

Hence, the range is:

      $y\geq -2$

5)

$k(x)=2-\sqrt{x}$

Since,

$\sqrt{x}\geq 0$

This means that:

$-\sqrt{x}\leq 0$

so,

$-\sqrt{x}+2\leq 2\\\\i.e.\\\\2-\sqrt{x}\leq 2\\\\i.e.\\\\k(x)\leq 2$

Hence, the range is:

                      $y\leq 2$

6)

$v(x)=\sqrt{-x}$

We know that the square root of a number is always greater than or equal to zero.

so,

$\sqrt{-x}\geq 0$

i.e.

The range is:

                  $y\geq 0$

Answer 2

$1.\ t(x)=-\sqrt{x}\to y\leq0\\\\2.\ p(x)=\sqrt[3]{2-x}\to \text{All real numbers}\\\\3.\ w(x)=2+\sqrt{x}\to y\geq2\\\\4.\ r(x)=-2+\sqrt{2-x}\to y\geq-2\\\\5.\ k(x)=2-\sqrt{x}\to y\leq2\\\\6.\ v(x)=\sqrt{-x}\to y\geq0\\-------------------\\\sqrt{x}\geq0,\ \sqrt[3]{x}\in\mathbb{R}$