A little inverse help cube root

[aisha]

find the inverse of f(x)=3sqrt(x+2) -7 x>=-2

I got this far (x+7)=3sqrt(y+2)

I don't know how to get rid of the cube root someone help please

 

[apchemstudent]

find the inverse of f(x)=3sqrt(x+2) -7 x>=-2

I got this far (x+7)=3sqrt(y+2)

I don't know how to get rid of the cube root someone help please

just cube both sides of the equation...

 

[aisha]

therefore is the inverse of my equation f^-1(x)=(x+7)^(3)-2 ? :yuck:

 

[Parth Dave]

Yup.....

 

[aisha]

it doesn't look anything like the inverse when plugged into the graphing calculator into y= and compared to DrawInv

 

[aisha]

I tried again now my answer is

[(x+7)^3] -2
_________
27

Is this correct? OR totally wrong? :uhh:

 

[Parth Dave]

Is that 3 * cube root of (x + 2)

or just cube root of (x + 2)??

 

[dextercioby]

find the inverse of f(x)=3sqrt(x+2) -7 x>=-2
I got this far (x+7)=3sqrt(y+2)
I don't know how to get rid of the cube root someone help please

This is really curious.She's posted 196 times,yet she hasn't had the time to learn how to edit the formulas in 'tex'.
If your initial function was:
$$f_{1}(x)=\sqrt[3]{x+2}-7$$
,then the inverse is found simply:
$$f_{1}^{-1}(x)=(x+7)^{3}-2$$.
If your initial function was:
$$f_{2}(x)=3\sqrt[3]{x+2}-7$$
,then the inverse is easily found to be
$$f_{2}^{-1}(x)=\frac{(x+7)^{3}}{27}-2$$
The curious part is that i don't have any idea "which is which",as you posted both answers as 'correct'.Obviously only one is.
In the end i'd like to ask you in a very polite way to sacrifice some of your spare time and read the .pdf document which explains you how you can edit your mathematical formulas in 'tex'.

Daniel.

PS.Isn't it a little weird to you as well that on Christmas Evening you chose mathematics as a 'date'?? :tongue2:

 

[HallsofIvy]

By the way- the notation "3 sqrt(x)" to mean "cube root" really steams me. It makes it look like you don't know the difference between "cube root" and "square root". Even if you refuse to use $^3\sqrt{x}$ or ³√(x), you could at least write "cbrt(x)" or "3rdrt(x)".

 

[aisha]

Ur right I really don't have time to read the txt, but when I do have I time I will for sure. The question is 3 * the square root of (x+2) subtract 7
(7 is not under the square root). And yes I do know the difference between cube root and square root.

I got the previous answer because I was told to cube both sides of the equation to get rid of the square root, but I think I was supposed to square both sides of the equation to get rid of the square root.

So is the answer
(x+7)^(2) -2
---------
9

 

[dextercioby]

Yes,this time it's the good answer. One request,though,pleeaeaeaeaeaeaeaeaeaeaeaeaeaeeaeaeaeaeeaeaeaeaeaeaseread that file which explains the tex editing language.U might be able to have an adjustable fraction line,not a bunch of minuses... :tongue2:

Daniel.

 

[BobG]

PS.Isn't it a little weird to you as well that on Christmas Evening you chose mathematics as a 'date'?? :tongue2:

What's wrong with that? I never get any dates, either. Of course, in my case, the reason is obvious.