Chapter 3.6, Problem 6MRPS

Solution

Two functions whose graphs are translations of the graph of the given function, such that the first function should have a domain of $x\ge 4$ and the second function should have a range of $y\ge -2$ .

It has been determined that the first function is $y=\sqrt{x-4}$ and the second function is $y=\sqrt{x}-2$ .

Given:

The function, $y=\sqrt{x}$ .

Concept used:

For real valued functions, the quantity under a square root and unless specified otherwise, the square root quantity must both be positive.

Calculation:

The given function is $y=\sqrt{x}$ .

Then, it must follow that $x\ge 0$ and $y\ge 0$ .

Now, it is given that the first function should have a domain of $x\ge 4$ .

That is, the first function should only be defined when $x-4\ge 0$ .

Now, the domain is $x\ge 0$ when the function is $y=\sqrt{x}$ .

Similarly, the domain is $x-4\ge 0$ when the function is $y=\sqrt{x-4}$ .

So, the first function is $y=\sqrt{x-4}$ . Note that this is a translation of the function $y=\sqrt{x}$ .

Now, it is given that the second function should have a range of $y\ge -2$ .

That is, for the second function, it must follow that $y+2\ge 0$ .

Now, the range is $y\ge 0$ when the function is $y=\sqrt{x}$ .

Similarly, the range is $y+2\ge 0$ when the function is $y+2=\sqrt{x}$ .

So, the second function is $y+2=\sqrt{x}$ or equivalently $y=\sqrt{x}-2$ . Note that this is a translation of the function $y=\sqrt{x}$ .

Conclusion:

It has been determined that the first function is $y=\sqrt{x-4}$ and the second function is $y=\sqrt{x}-2$ .