Finding the inverse of a function?

[nukeman]

Finding the inverse of a function?

Homework Statement

Find (f^-1)'(a), a =2

√(x^3 + x^2 +x +1)

So, if a = 2, then f^-1(2) = 1 and f(1) = 2

Homework Equations

The Attempt at a Solution

I figured out that f(1) = 2,

so

√(3(1)^2 + 2(1) + 1)

= √6

so the final answer I got was 1/√(6)

?

 

[Mark44]

Homework Statement

Find (f^-1)'(a), a =2

√(x^3 + x^2 +x +1)

So, if a = 2, then f^-1(2) = 1 and f(1) = 2

Homework Equations

The Attempt at a Solution

I figured out that f(1) = 2,

so

√(3(1)^2 + 2(1) + 1)

= √6

so the final answer I got was 1/√(6)

?

They are not asking for f^-1(2) -- you need to get (f^-1)'(2).

Start by differentiating each side with respect to x to eventually end up with y = (f^-1)'(x).

 

[nukeman]

I don't understand. What is the first thing I do?

Would it be to figure out what value of f(x) would equal 2?

In this case, 1 would correct?

 

[Mark44]

I don't understand. What is the first thing I do?

Would it be to figure out what value of f(x) would equal 2?

In this case, 1 would correct?

I don't think this is relevant in this problem. They are asking you about the derivative of the inverse, not the inverse.

 

[gopher_p]

I don't think this is relevant in this problem. They are asking you about the derivative of the inverse, not the inverse.

He's trying to use [itex](f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}[/itex].

 

[nukeman]

Correct, I am

 

[gopher_p]

Correct, I am

So then [itex](f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(1)}=\ldots[/itex]

Find [itex]f'(1)[/itex] and you're done.

Edit: I now see where you tried to do this. Maybe take your time with that derivative. You need to use the power rule and the chain rule.

P.S. I reckon [itex](f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}[/itex] is a fairly relevant equation here. Probably should have put that in the first post. You can't assume that tutors remember everything from their undergrad (maybe even high school) calc course.

 

[nukeman]

I don't understand how the answer is 2/3 :(

 

[gopher_p]

Three questions:

1) What is the derivative (with respect to [itex]x[/itex]) of [itex]\sqrt{x}[/itex]? (hint: you need the power rule)

2) Assuming [itex]u[/itex] is a differentiable function of [itex]x[/itex], what is the derivative with respect to [itex]x[/itex] of [itex]\sqrt{u}[/itex]? (hint: you need the chain rule)

3) What is the derivative of [itex]\sqrt{x^3 + x^2 +x +1}[/itex]? (hint: let [itex]u=x^3 + x^2 +x +1[/itex] and use part 2)

 

[Dick]

I don't understand how the answer is 2/3 :(

Maybe that's because you haven't worked out what f' is. Do that and then figure out what 1/f'(1) is.