Inverse problem?

- AmTran_Bus

Inverse problem?

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- AmTran_Bus

I am willing to work it out, just need help.

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- AmTran_Bus

The inverse is y=x^2+4, right?

- geerky42

What is "Theorem"?

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- geerky42

Correct

- AmTran_Bus

Hum. Well, this is online, but let me reference a book.

- AmTran_Bus

Maybe this?|dw:1433387173701:dw|

- geerky42

Looks right, but seem useless, since we can just use g(x) to derivative with.

- geerky42

We know \[g(x) =f^{-1}(x)= x^2+4\]

- AmTran_Bus

Right

- geerky42

well, you should be able to handle the rest.

- AmTran_Bus

I mean, do you just plug it in?

- geerky42

well, take derivative of g(x), then plug in

- AmTran_Bus

2x, or 8...

- geerky42

\(g(x) = x^2+4\) So \(g'(x) = 2x\)
Now plug in \(x=4\);
\[g'(4) = \boxed{8}\]

- AmTran_Bus

Yay!

- AmTran_Bus

Thank you

- geerky42

No problem

- AmTran_Bus

Hey @geerky42, would I do this the exact same way?

##### 1 Attachment

- AmTran_Bus

Its inverse is y= 2+x^2+tan(pi x/2)

- geerky42

That's exactly same function as original function.

- geerky42

its inverse should have arctan

- AmTran_Bus

My bad.

- AmTran_Bus

Can you help me continue on then? I haven't really studied that in this context.

- geerky42

Did you find inverse? You can swap x and y to each other, then isolate tan(...)

- geerky42

from there, take arctan.

- AmTran_Bus

Eek man, why is this one so hard?

- geerky42

Gee didn't realize that finding inverse would be hard lol...

- geerky42

@ganeshie8 Can you help?

- ganeshie8

I don't seem to remember the inverse derivative formula... deriving it is easy
we knw that \[\large f(f^{-1}(x)) = x\]
differentiating both sides gives
\[\large f'(f^{-1}(x)){f^{-1}}'(x)=1 \]
\[\large {f^{-1}}'(x) = \dfrac{1}{ f'(f^{-1}(x))}\]

- ganeshie8

plugin \(x=2\)

- geerky42

Hmm.
What is \(f(0)\)? @AmTran_Bus

- geerky42

It seems that we need to use reasoning here.

- AmTran_Bus

I think we are maybe overthinking. Look at the choices they give as answers.

- ganeshie8

can you find f(0) ?

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