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Mathematics 18 Online
KyledaGreat:

Find a formula for the inverse of the following function, if possible. or does not have an inverse function

KyledaGreat:

\[h(x) = \sqrt[3]{x^3+ 3} + 3\] \[h^{-1}(x) =\]

KyledaGreat:

@Tranquility

KyledaGreat:

(x^3/3)

KyledaGreat:

@tranquility

Tranquility:

@kyledagreat wrote:
(x^3/3)
That's not quite right

Tranquility:

@kyledagreat wrote:
\(h(x) = \sqrt[3]{x^3+ 3} + 3\) \(h^{-1}(x) =\)
replace x with \( h^{-1}(x)\) or just y and the h(x) with x \(x = \sqrt[3]{y^3+ 3} + 3\)

Tranquility:

All you need to do is solve for y now What do you get when you take the 3 to the other side?

KyledaGreat:

d\[y = \sqrt[3]{x^3 - 9x^2 + 27x - 30}\]

Tranquility:

No

Tranquility:

I'm not sure how you got that inside the cube root

Tranquility:

Oh wait, you're saying that as your final answer mhmm Let me check, I haven't even gotten that far yet

Tranquility:

\(x = \sqrt[3]{y^3+ 3} + 3\) (x - 3)^3 - 3 = y^3 y = cube root of all that so it looks correct

Tranquility:

Yeah that's correct for the inverse

KyledaGreat:

alright , which one should i enter to be correct ?

Tranquility:

I would think that the program should accept both answers

KyledaGreat:

no it didn't , i have to try another one . Could you check and see the other one

Tranquility:

cube root of (x - 3)^3 - 3

KyledaGreat:

oh ok , i posted a new one

Tranquility:

sure

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