# Thread: Inverse and differentation of an equation

1. How do I solve this. Getting an inverse alone seems to be a quite long route. Is there an easier way of doing this?  Reply With Quote

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3. Originally Posted by ertagon2 How do I solve this. Getting an inverse alone seems to be a quite long route. Is there an easier way of doing this?
edit: Please read and respond to post #3 first (it provides better help) and then check the hidden content here to confirm your result from the indicated differentiation.   Reply With Quote

4. As an aside, do you know why it is an injective function?

Because of injectivity, we can regard $x$ as a function of $y$, for $y$ in the range of $f$.
Is the point $y = -3$ in the range of $f$?
Could you try to differentiate the identity $y = f(x(y))$ with respect to $y$?

(The last step will reproduce the formula that was given in post #2.)  Reply With Quote Originally Posted by MarkFL edit: Please read and respond to post #3 first (it provides better help) and then check the hidden content here to confirm your result from the indicated differentiation. We did cover the formula's derviation in lectures, but how do I use. Do I still have to get the inverse of f(x)?  Reply With Quote

6. Originally Posted by ertagon2 We did cover the formula's derviation in lectures, but how do I use. Do I still have to get the inverse of f(x)?
We need to find $f^{-1}(-3)$, but we don't have to find $f^{-1}$ explicitly. If we observe we must have:

$\displaystyle f\left(f^{-1}(-3)\right)=-3$

So, let's define:

$\displaystyle b=f^{-1}(-3)$

And write:

$\displaystyle f(b)=-3$

Can you solve for $b$?  Reply With Quote Originally Posted by MarkFL We need to find $f^{-1}(-3)$, but we don't have to find $f^{-1}$ explicitly. If we observe we must have:

$\displaystyle f\left(f^{-1}(-3)\right)=-3$

So, let's define:

$\displaystyle b=f^{-1}(-3)$

And write:

$\displaystyle f(b)=-3$

Can you solve for $b$?
Nope, I see that you did in the first 2 lines but what do you mean by solving for b?  Reply With Quote

8. Originally Posted by ertagon2 Nope, I see that you did in the first 2 lines but what do you mean by solving for b?
$f(b)=2b^3+7b-3$

And so you want to solve:

$\displaystyle 2b^3+7b-3=-3$

for its real root.   Reply With Quote

9. I would like to return to the question posted by Krylov: Originally Posted by Krylov As an aside, do you know why it is an injective function?...
Your professor felt it important for you to be able to discuss why we know this function must be injective, and I also agree this is an important point when discussing inverse functions. Can you show why the given function must be one-to-one?  Reply With Quote

10. Originally Posted by MarkFL Your professor felt it important for you to be able to discuss why we know this function must be injective, and I also agree this is an important point when discussing inverse functions. Can you show why the given function must be one-to-one?
Good post. If I may add without giving the answer away; think of what you are doing when you find the inverse by switching variables and solving.

Without giving anymore details, think of the vertical line test and horizontal line test. The result of one-one should follow. Just trying to help with some graphical representation.

Think about the function $f(x)= x^2$
What happens at the points $x= (-1)$ and $(1)$?

So we have a function, that is what we know; now try to find the inverse if it has one. Then try to deduce the importance on the function being one to one.  Reply With Quote Originally Posted by MarkFL Your professor felt it important for you to be able to discuss why we know this function must be injective, and I also agree this is an important point when discussing inverse functions. Can you show why the given function must be one-to-one?
These are notes from the lecture that covered this topic.
He only showed us what an injective function is and derived the formula. I honestly have no idea how to attempt this question.  Reply With Quote

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