Inverted function

[pope4]

The question asks to write h(x)=-(x+3)^2-5 inverted and explain what restrictions to put on the inversion's domain to make it a function. The answer came out to be

\(\displaystyle y=\pm \sqrt{-x-5}\) - 3,

and it restricts x from being less than -3. Why is there a plus minus sign at the beginning? And if x is equal to anything more than -5, doesn't that make the square root impossible since you can't square root a negative?

 

[Dr.Peterson]

The question asks to write h(x)=-(x+3)^2-5 inverted and explain what restrictions to put on the inversion's domain to make it a function. The answer came out to be

\(\displaystyle y=\pm \sqrt{-x-5}\) - 3,

and it restricts x from being less than -3. Why is there a plus minus sign at the beginning? And if x is equal to anything more than -5, doesn't that make the square root impossible since you can't square root a negative?

Can you show us the exact problem and answer, perhaps as images, so we can be sure exactly what they are saying? There are a couple oddities here, one being that the answer, an inverse function, should be a function definition, not an equation containing y; another is that it is x in the original function, h, not in the inverse function, that must be restricted (that is, in the domain of h, not of the inverse as you say). Yet another oddity is that there are two different answers, based on different restrictions of the original domain. (The plus-or-minus is probably intended to combine the two possible answers, or else to show the inverse relation before you make it a function.)

Once these issues are resolved, there is a lot to be said in explanation. Here is one explanation I found for just this sort of problem:

Sosmath.com

www.sosmath.com

 

[pope4]

Can you show us the exact problem and answer, perhaps as images, so we can be sure exactly what they are saying? There are a couple oddities here, one being that the answer, an inverse function, should be a function definition, not an equation containing y; another is that it is x in the original function, h, not in the inverse function, that must be restricted (that is, in the domain of h, not of the inverse as you say). Yet another oddity is that there are two different answers, based on different restrictions of the original domain. (The plus-or-minus is probably intended to combine the two possible answers, or else to show the inverse relation before you make it a function.)

Once these issues are resolved, there is a lot to be said in explanation. Here is one explanation I found for just this sort of problem:

Sosmath.com

www.sosmath.com

Here are screenshots of the question and the answer according to the textbook!

 

[Deleted member 4993]

Why is there a plus minus sign at the beginning?

If

x² = 4 ...........→............. x = ±2

 

[Dr.Peterson]

Here are screenshots of the question and the answer according to the textbook!View attachment 34015View attachment 34016

Thanks. That helps a lot.

​

What you showed, \(\displaystyle y=\pm \sqrt{-x-5}-3\), is, just as (b) asks, not an inverse function, but an inverse relation -- it has two values for many inputs.

Now observe that (c) doesn't ask, as you said, "what restrictions to put on the inversion's domain to make it a function", but what restrictions could be put on the [original] function's domain so that the inverse is a function. It's important to read (and quote) carefully.

The most important word for our purposes is could. The answer you quoted, "it restricts x from being less than -3", however I interpret it, is only one possible answer. You can either restrict the domain to [imath]x\ge-3[/imath] or [imath]x\le-3[/imath]; either of those will make it a function. This corresponds to whether you choose the positive or negative sign in the relation, in order to give it only one value and make it a function.

And that's why their answer says

​

The reference I gave explains all this. Be sure to read it, and/or your textbook.

 

[pope4]

Thanks. That helps a lot.

View attachment 34017​

What you showed, \(\displaystyle y=\pm \sqrt{-x-5}-3\), is, just as (b) asks, not an inverse function, but an inverse relation -- it has two values for many inputs.

Now observe that (c) doesn't ask, as you said, "what restrictions to put on the inversion's domain to make it a function", but what restrictions could be put on the [original] function's domain so that the inverse is a function. It's important to read (and quote) carefully.

The most important word for our purposes is could. The answer you quoted, "it restricts x from being less than -3", however I interpret it, is only one possible answer. You can either restrict the domain to [imath]x\ge-3[/imath] or [imath]x\le-3[/imath]; either of those will make it a function. This corresponds to whether you choose the positive or negative sign in the relation, in order to give it only one value and make it a function.

And that's why their answer says

View attachment 34018​

The reference I gave explains all this. Be sure to read it, and/or your textbook.

Thank you! The reference explained it all, I really appreciate your help!