Hard Inverse function f(x) problem

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wolly

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So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
 
wolly said:
So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
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Would you be able to invert y = A*x^2 + B*x + C?
 
wolly said:
So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
Click to expand...

You should derive it at least once in your mathematics career.

\(\displaystyle f(h(x)) = x\) - Definition of Inverse

\(\displaystyle \dfrac{d}{dx}\;f(h(x)) = \dfrac{d}{dx}\;x\) - Introduce the Derivative

\(\displaystyle f'(h(x))\cdot h'(x) = 1\) - Execute the Derivative (Chain Rule)

Solve for h'(x).

You would have to be more careful with g(x) = x^3 - x + 1. Do you know why?
 
wolly said:
So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
Click to expand...
(f^-1)(3)\(\displaystyle \neq\)5 since f(5)\(\displaystyle \neq\)3. Note that f(5) = 131
 
wolly said:
So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
Click to expand...
Answered at my math forum
 
wolly said:
So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
Click to expand...

Please state the exact words of the problem you were given. I suspect that you are trying to do more than was asked.

If it was just to find f-1(3), I would do that by solving the equation f(x) = 3, which can be done by trial and error as well as anything.

Also, can you quote what was said about using differentiation to find an inverse?
 
Dr.Peterson said:
Please state the exact words of the problem you were given. I suspect that you are trying to do more than was asked.

If it was just to find f-1(3), I would do that by solving the equation f(x) = 3, which can be done by trial and error as well as anything.

Also, can you quote what was said about using differentiation to find an inverse?
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Ok.I have this function and I want to know how I can calculate the inverse function from the original function.I know that f(1)=3 but how does that help me?Do I equal f(x) with 3?
 
wolly said:
Ok.I have this function and I want to know how I can calculate the inverse function from the original function.I know that f(1)=3 but how does that help me?Do I equal f(x) with 3?
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You have been asked to post the exact wording. So this a guess.
Have you been asked to find the value of the derivative of the inverse function at \(\displaystyle x=3~?\) i.e. \(\displaystyle (f^{-1})'(3)\)
 
Ok so I know that f(x)=x^3+x+1 and if I plug in the value f(1)=3 I get 3.I guess that if I change variables from x=3 and y=1 to y=3 and x=1 I get f(^(-1))'(3)=1.This is what I want to proove.If x=1 and y=3 does it mean that the function f(x) is inverse?I'm asking this because I don't know what other proof could help me to get f(^(-1))'(3)=1.Is my little proof correct?
 
wolly said:
So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
Click to expand...

wolly said:
Ok.I have this function and I want to know how I can calculate the inverse function from the original function.I know that f(1)=3 but how does that help me?Do I equal f(x) with 3?
Click to expand...

wolly said:
Ok so I know that f(x)=x^3+x+1 and if I plug in the value f(1)=3 I get 3.I guess that if I change variables from x=3 and y=1 to y=3 and x=1 I get f(^(-1))'(3)=1.This is what I want to proove.If x=1 and y=3 does it mean that the function f(x) is inverse?I'm asking this because I don't know what other proof could help me to get f(^(-1))'(3)=1.Is my little proof correct?
Click to expand...

Please state the exact wording of the problem you are trying to answer. It is very important!

If you were asked to find the inverse function (that is, a formula for all x), it probably can't be done (or rather, is very complicated).

If you were asked only to find f^-1(3), when all you need to do is to interchange the variables:

f(x) = y is equivalent to f^-1(y) = x.

So, in particular,

f(1) = 3 is equivalent to f^-1(3) = 1.

But you also mentioned differentiation, leading me to wonder it you were really supposed to find the derivative of the inverse function at x=3, though you never said that. It sounded like you wanted to use differentiation to find the inverse, which doesn't make much sense. What you write now looks like you want the derivative of the inverse, but you haven't shown any relevant work.

If so, you have been told how the derivative of the inverse is related to the derivative of f. What is that relationship? Have you tried doing that?
 
wolly said:
Ok so I know that f(x)=x^3+x+1 and if I plug in the value f(1)=3 I get 3.I guess that if I change variables from x=3 and y=1 to y=3 and x=1 I get f(^(-1))'(3)=1.This is what I want to proove.If x=1 and y=3 does it mean that the function f(x) is inverse?I'm asking this because I don't know what other proof could help me to get f(^(-1))'(3)=1.Is my little proof correct?
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See whether the fifth post in this thread answers your question.

http://mymathforum.com/pre-calculus/345432-hard-inverse-function-f-x-problem.html

Please, in the future, give the problem exactly and completely as given to you
 
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wolly said:
Ok so I know that f(x)=x^3+x+1 and if I plug in the value f(1)=3 I get 3.I guess that if I change variables from x=3 and y=1 to y=3 and x=1 I get f(^(-1))'(3)=1.This is what I want to proove.If x=1 and y=3 does it mean that the function f(x) is inverse?I'm asking this because I don't know what other proof could help me to get f(^(-1))'(3)=1.Is my little proof correct?
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To wolly, you were repeatedly asked to supply the exact wording of the question. Had you done, this thread could have shorten considerably,
Asking for the derivative of the inverse does not require finding the inverse.
You were given that \(\displaystyle f(x_0)=y_0\) and asked to find \(\displaystyle D_x[f^{-1}(y_0)]\).

We know that \(\displaystyle f^{-1}(f(x))=x\)
\(\displaystyle \begin{align*}f(f^{-1})(x)&=x\text{ so that} \\D_x[(f^{-1}(f(x_0))]&=D_x[x]\\(f^{-1})'(y_0)f'(x_0)&=1\end{align*}\)

Thus \(\displaystyle (f^{-1}(3))'=\dfrac{1}{?}\)
 
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wolly said:
Ok so I know that f(x)=x^3+x+1 and if I plug in the value f(1)=3 I get 3.I guess that if I change variables from x=3 and y=1 to y=3 and x=1 I get f(^(-1))'(3)=1.This is what I want to proove.If x=1 and y=3 does it mean that the function f(x) is inverse?I'm asking this because I don't know what other proof could help me to get f(^(-1))'(3)=1.Is my little proof correct?
Click to expand...
If f(a) = b, then f-1(b) = a. NOT (f-1)' (b) = a
 
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