Quick Question: Let f(x) = 6+7x+2cosx. Find(f^-1)'(8)

[boardr]

I'm stuck on a problem, any help is appreciated. I know the derivative is 7-2sinx and I know the equation is (f^-1)'(x)=1/f'(f^-1(x)), just confused a bit.

Let f(x) = 6+7x+2cosx Find(f^-1)'(8)

 

[Dr.Peterson]

I'm stuck on a problem, any help is appreciated. I know the derivative is 7-2sinx and I know the equation is (f^-1)'(x)=1/f'(f^-1(x)), just confused a bit.

Let f(x) = 6+7x+2cosx Find(f^-1)'(8)

A key step is to find f^-1(8). (Don't try to find it algebraically; guess!)

 

[Dr.Peterson]

Please show your work, so we can see where you are confused. I don't yet know whether my hint is what you needed.

 

[boardr]

So I know f(0)=8, do I put 0 into f'? So it's 1/f'(0)?

 

[pka]

I'm stuck on a problem, any help is appreciated. I know the derivative is 7-2sinx and I know the equation is (f^-1)'(x)=1/f'(f^-1(x)), just confused a bit. Let f(x) = 6+7x+2cosx Find(f^-1)'(8)

Because you have shown us none of your thinking, despite your being asked, it is hard to know what you fail to understand.
\(\displaystyle f(x) = 6+7x+2\cos(x)\), do you know what \(\displaystyle f(0)=~?\)
Do you know what it is that you are being required to do? If you do, tell us in some detail.
If not, the ask some questions about process.

 

[boardr]

Because you have shown us none of your thinking, despite your being asked, it is hard to know what you fail to understand.
\(\displaystyle f(x) = 6+7x+2\cos(x)\), do you know what \(\displaystyle f(0)=~?\)
Do you know what it is that you are being required to do? If you do, tell us in some detail.
If not, the ask some questions about process.

I get f(0)= 8. I don't understand how that helps in this situation. It is asking for the f inverse prime of 8. If it just asked for the f inverse I would be able to do it, I just haven't had a problem like this before.

 

[boardr]

Also I did reply, I'm new to the forum though so it has to go through a moderator before the post shows up.

 

[Dr.Peterson]

So I know f(0)=8, do I put 0 into f'? So it's 1/f'(0)?

I get f(0)= 8. I don't understand how that helps in this situation. It is asking for the f inverse prime of 8. If it just asked for the f inverse I would be able to do it, I just haven't had a problem like this before.

Now that you have written several posts, you won't be moderated, and we won't have this frustrating delay!

You are right that, since f(0_ = 8, f^-1(8) = 0. So the answer is, indeed, 1/f'(0), as you said. You evidently do understand how it helps!

So, what do you get for the answer?

You might then graph f, and think about how the graph shows the result you found.

 

[boardr]

So it is 1/7? Originally I had put 1/(7pi/6), that's where my confusion came from. Thanks for the help, I'll try to be more clear in the future.

 

[Dr.Peterson]

So it is 1/7? Originally I had put 1/(7pi/6), that's where my confusion came from. Thanks for the help, I'll try to be more clear in the future.

Can you tell me why you thought it was 1/(7pi/6)? Maybe there's a misunderstanding we need to discuss.

But, yes, the slope of f^-1 at (8,0) is 1/7, because the slope of f at (0,8) is 7.

 

[boardr]

I was thinking I had to make f' = 8 instead of f, where f(0) =8, f'(7pi/6)=8. I just flipped something in my mind I guess.

 

[Dr.Peterson]

I was thinking I had to make f' = 8 instead of f, where f(0) =8, f'(7pi/6)=8. I just flipped something in my mind I guess.

Ah! So you used (f ')^-1(8) instead of f '(f^-1(8)) . Yes, there are a lot of pieces to keep track of here!

 

[pka]

I was thinking I had to make f' = 8 instead of f, where f(0) =8, f'(7pi/6)=8. I just flipped something in my mind I guess.

Lets take a tour through the derivation.
Recall the derivative of a composition. Suppose that \(\displaystyle f~\&~g\) are both differentiable then \(\displaystyle D_x[f\circ g(x)]=D_x[f(g(x))=f'(g(x))\cdot g'(x)\)
If that is not absolutely clear to you, the stop and go back and review derivatives of composition.

Next review the idea of inverse function \(\displaystyle f^{-1}(x)\) is such \(\displaystyle f\circ f^{-1}(x)=f(f^{-1}(x))=x\).
So using the derivative of that composition:
\(\displaystyle \begin{align*}D_x[f(f^{-1}(x))]&=D_x[x]\\f'(f^{-1}(x))\cdot[f^{-1}(x)]'&=1 \\ [f^{-1}(x))]'&=\dfrac{1}{f'(f^{-1}(x))}\end{align*}\)

 

[Dr.Peterson]

I'm stuck on a problem, any help is appreciated. I know the derivative is 7-2sinx and I know the equation is (f^-1)'(x)=1/f'(f^-1(x)), just confused a bit.

Let f(x) = 6+7x+2cosx Find(f^-1)'(8)

\(\displaystyle \begin{align*}D_x[f(f^{-1}(x))]&=D_x[x]\\f'(f^{-1}(x))\cdot[f^{-1}(x)]'&=1 \\ [f^{-1}(x))]'&=\dfrac{1}{f'(f^{-1}(x))}\end{align*}\)

In the original post, boardr stated this formula. The issue is how to apply it correctly. I think that's been worked out.

 

[pka]

In the original post, boardr stated this formula.

I am fully aware that boardr says that s/he knows the 'formula'.
But many years experience taught me that can be useless.
The same experience taught me that if a student understands the derivation then its correct application is more likely.