Inverse Function

[thetguy]

If (3, 5) is an element of f, and f'(3) = (2/3), what is (f^-1(5))'?

I'm almost positive this is an easy question, but don't remember how inverse functions work.

 

[MarkFL]

Hello, and welcome to FMH!

Recall that:

[MATH]f^{-1}(f(x))=x[/MATH]
And we are told:

[MATH]f(3)=5[/MATH]

 

[Dr.Peterson]

If (3, 5) is an element of f, and f'(3) = (2/3), what is (f^-1(5))'?

I'm almost positive this is an easy question, but don't remember how inverse functions work.

An inverse function swaps the roles of input and output.

If (a,b) is on the graph of a function f, that means that f(a) = b.

Then (b,a) is on the graph of the inverse function f^-1; that is, f^-1(b) = a.

So you're just reading backward to get the inverse. f(a) = b means the same thing as f^-1(b) = a.

 

[thetguy]

So f^-1(5) would be 3, in this case I believe. So according to the problem, would that mean, (f^-1(5))' = (3)' = 0, or am I missing something?

Edit: I guess what's confusing me is why the problem mentions f'(3) = (2/3). So should the answer be (f^-1(5))' = (3)' = 0, or (f^-1(5))' = (f(3))' = (2/3)?

 

[MarkFL]

I missed that you were asked to find:

[MATH][f^{-1}]'(5)[/MATH]
You should be familiar with the formula:

[MATH][f^{-1}]'(a)=\frac{1}{f'(f^{-1}(a))}[/MATH]
What do you then conclude?

 

[Dr.Peterson]

So f^-1(5) would be 3, in this case I believe. So according to the problem, would that mean, (f^-1(5))' = (3)' = 0, or am I missing something?

Edit: I guess what's confusing me is why the problem mentions f'(3) = (2/3). So should the answer be (f^-1(5))' = (3)' = 0, or (f^-1(5))' = (f(3))' = (2/3)?

I missed that part of the question too; in part that's because your notation is atrocious. (I hope that isn't what the problem itself says!) I took your question to be just about the inverse itself.

Since [MATH]f^{-1}(5) = 3[/MATH] (just a number), literally [MATH][f^{-1}(5)]'[/MATH] would be zero, as you said! What you meant, as MarkFL correctly wrote, is [MATH][f^{-1}]'(5)[/MATH]. That means the derivative of the inverse function, applied to the input 5, not the derivative of the inverse function of 5. Both inverse and derivative operate on functions themselves, not their values.

 

[thetguy]

Ah, I see. We were not introduced to that formula in class yet. Our calculus class was moved online, and due to the transition, some information has been lost in translation.

So according to the formula:

[f^-1]'(5) = 1/(f'(f^-1(5)) = 1/(f'(3)) = 1/(2/3) = (3/2)

Do I follow this correctly?

Edit: I understand the notation is awful, but this is exactly how our professor wrote it on the board. So when I posted it here, I posted it exactly as written.

 

[MarkFL]

Ah, I see. We were not introduced to that formula in class yet. Our calculus class was moved online, and due to the transition, some information has been lost in translation.

So according to the formula:

[f^-1]'(5) = 1/(f'(f^-1(5)) = 1/(f'(3)) = 1/(2/3) = (3/2)

Do I follow this correctly?

Edit: I understand the notation is awful, but this is exactly how our professor wrote it on the board. So when I posted it here, I posted it exactly as written.

Yes, I arrived at the same value.

 

[thetguy]

@MarkFL thank you for your help, and for the formula!

And @Dr.Peterson, thank you for the definition and clarification. That really helped clear up some confusion I had about derivative notation in general.

 

[pka]

Ah, I see. We were not introduced to that formula in class yet. Our calculus class was moved online, and due to the transition, some information has been lost in translation.
So according to the formula:
[f^-1]'(5) = 1/(f'(f^-1(5)) = 1/(f'(3)) = 1/(2/3) = (3/2)
Do I follow this correctly?

Using the notation for function composition here is a formal proof.
$f^{-1}\circ f(x)=x$
\(\begin{align*}\dfrac{d}{dx}\left((f^{-1}\circ f(x)\right)&=\dfrac{d}{dx}(x) \\\left(f^{-1}\right)^{\prime}(f(x)) f^{\prime}(x)&=1\\\left(f^{-1}\right)^{\prime}\circ(f(x))(f(x))&=\dfrac{1}{f^{\prime}(x)} \end{align*}\)

 

[Steven G]

I understand the notation is awful, but this is exactly how our professor wrote it on the board. So when I posted it here, I posted it exactly as written.

If that is the case then:
Assuming f(x) and its inverse are real valued functions defined for values we are concerned with, then f^-1(5) is just a number (in this case 3) and its derivative is 0. Any other answer is wrong.

 

[pka]

Using the notation for function composition here is a formal proof.
$f^{-1}\circ f(x)=x$
\(\begin{align*}\dfrac{d}{dx}\left((f^{-1}\circ f(x)\right)&=\dfrac{d}{dx}(x) \\\left(f^{-1}\right)^{\prime}(f(x)) f^{\prime}(x)&=1\\\left(f^{-1}\right)^{\prime}\circ(f(x))&=\dfrac{1}{f^{\prime}(x)} \end{align*}\)