Question

Match each function with its range.

1. t(x) = -sqrt of x                    |         y <= 2
2. p(x) = cube root of 2 - x      |         y <= 0
3. w(x) = 2 + sqrt of x             |         All real numbers
4. r(x) = -2 + sqrt of 2 - x        |        y >= 2
5. k(x) = 2 - sqrt of x               |        y >= -2
6. v(x) = sqrt of -x                   |        y >= 0

Answer 1

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

Answer 2

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

Explanation:

1)

`t(x)=-√(x)`

We know that:

`√(x)\geq 0`

This means that:

`-√(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+√(x)`

Again we know that:

`√(x)\geq 0\\\\i.e.\\\\2+√(x)\geq 2\\\\i.e.\\\\w(x)\geq 2`

Hence, the range is:

`y\geq 2`

4)

`r(x)=-2+√(2-x)`

we know that:

`√(x)\geq 0\\\\i.e.\\\\-2+√(x)\geq -2\\\\i.e.\\\\w(x)\geq -2`

Hence, the range is:

`y\geq -2`

5)

`k(x)=2-√(x)`

Since,

`√(x)\geq 0`

This means that:

`-√(x)\leq 0`

so,

`-√(x)+2\leq 2\\\\i.e.\\\\2-√(x)\leq 2\\\\i.e.\\\\k(x)\leq 2`

Hence, the range is:

`y\leq 2`

6)

`v(x)=√(-x)`

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

so,

`√(-x)\geq 0`

i.e.

The range is:

`y\geq 0`