Question

This question is complex to write so I hope everyone understands.

Find f^-1(x). f(x) = Cube Root 3\/x-9

Answer 1

solve
\[x=\sqrt[3]{y-9}\] for \(y\)

Answer 2

cube both sides to get
\[x^3=y-9\] then add 9 to both sides and you are done

Answer 3

y = x^3+9

Answer 4

Is that:\[y=\sqrt[3]{x-9}\]Since the inverse has the range of the original function as its domain and the domain of the original function as its range, we switch the y and x:\[\rightarrow x = \sqrt[3]{y-9}\] Now you have to solve for y and get the equation in the form, y = ....

Can you do that?

@MixMasterTX

Answer 5

And yes that's correct. Good job. @MixMasterTX

Answer 6

So when you cube both sides it cancels out the cube root on the right side?

Answer 7

Yup.

Answer 8

Taking both sides to the power/exponent of 3 makes the cube root go away. Then you can solve for y quite easily.

Answer 9

ok...so the f^-1(x) really plays no part in the equation?

Answer 10

Yes it does. Didn't you notice right at the beginning that Satellite formed the inverse by switching the x and the y?

Answer 11

Saying f(x) = something is the same as saying y = something. It just means that f is a function of x or y is changing with respect to x. So that way we can replace f(x) with y when we are finding inverses so it's easier and makes more sense when you graph it on the xy plane.

@MixMasterTX

Answer 12

ok...so by switching the x and the y it allows you to cube both sides thus eliminating the cube on the right side, adding 9 to both sides and solving.
y = x^3+9......Right?

Answer 13

Thanks to
@genius12
@satellite73
@Mertsj

Answer 14

No. By switching the x and the y, you are forming the inverse.