MHB Inverse and differentation of an equation

[ertagon2]

View attachment 7537
How do I solve this. Getting an inverse alone seems to be a quite long route. Is there an easier way of doing this?

 

[MarkFL]

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. :)

Spoiler

I would use:

$$\left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

Can you proceed?

 

[S.G. Janssens]

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.)

 

[ertagon2]

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. :)

Spoiler

I would use:

$$\left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

Can you proceed?

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)?

 

[MarkFL]

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:

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

So, let's define:

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

And write:

$$f(b)=-3$$

Can you solve for $b$?

 

[ertagon2]

We need to find $f^{-1}(-3)$, but we don't have to find $f^{-1}$ explicitly. If we observe we must have:

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

So, let's define:

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

And write:

$$f(b)=-3$$

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?

 

[MarkFL]

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:

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

for its real root. :)

 

[MarkFL]

I would like to return to the question 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?

 

[DrWahoo]

I would like to return to the question posted by Krylov:
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.

 

[ertagon2]

I would like to return to the question posted by Krylov:
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. http://hamilton.nuigalway.ie/teachingWeb/MA180_2017_18/calc_17.pdf
He only showed us what an injective function is and derived the formula. I honestly have no idea how to attempt this question.

 

[DrWahoo]

A function says that for every x, there is exactly one y. That is, y values can be duplicated but x values can not be repeated.

If the function has an inverse that is also a function, then there can only be one y for every x.

A one-to-one function, is a function in which for every x there is exactly one y and for every y, there is exactly one x. A one-to-one function has an inverse that is also a function.

So in our case where $f(x)=x^2$, the function obviously passes the vertical line test, but does not pass the horiztonal line test which implies it has no inverse. That is because when you choose any value such; ie $x=2$ and $x=-2$ we get an output of $4$. This should help with the injection/ one-to-one understanding.

So when we take the inverse of that function, these values change to $y=2$ , $y=-2$ when x=4. Is this a function? Why not?

 

[DrWahoo]

These are notes from the lecture that covered this topic. http://hamilton.nuigalway.ie/teachingWeb/MA180_2017_18/calc_17.pdf
He only showed us what an injective function is and derived the formula. I honestly have no idea how to attempt this question.

Look at this pdf and try to understand the reliance on one-to-one.

View attachment 7541

 

[MarkFL]

These are notes from the lecture that covered this topic. http://hamilton.nuigalway.ie/teachingWeb/MA180_2017_18/calc_17.pdf
He only showed us what an injective function is and derived the formula. I honestly have no idea how to attempt this question.

What I had in mind, is to look at the function's first derivative. If this derivative never changes sign, that is, has no real roots, or no real roots of odd multiplicity, then it can never change direction as it will never go from positive to negative or negative to positive. Functions that are strictly increasing or decreasing are one-to-one.

Does the sign of the given function's first derivative ever change?

 

[ertagon2]

I figured it out.

View attachment 7545

 

[HallsofIvy]

It is interesting that you want to solve f(x)= -3 and immediately jump to "f(0)= 2(0^3)+ 7(0)- 3= 3". How did you arrive at x= 0 to begin with?

 

[MarkFL]

I would have first dealt with the question regarding why the given function is injective. We find:

$$f'(x)=6x^2+7$$

Since $f'>0\,\forall\,x\in\mathbb{R}$, we know the function is strictly increasing, which means it is injective, or one-to-one.

Next, I would implicitly differentiate the given function with respect to $f$ to get:

$$1=\left(6x^2+7\right)\d{x}{f}\implies \d{x}{f}=\frac{1}{6x^2+7}$$

Next, we solve:

$$f(x)=-3$$

$$2x^3+7x-3=-3$$

$$x\left(2x^2+7\right)=0$$

The only real root is:

$$x=0$$

Thus, as the point $(0,-3)$ is on $f$, then $(-3,0)$ must be on $f^{-1}$ and so we may write:

$$\left[f^{-1}\right]'(-3)=\left.\d{x}{f}\right|_{x=0}=\frac{1}{7}\quad\checkmark$$

 

[HallsofIvy]

To show that a function is "injective" you have to show that the definition of "injective", that "if f(x)= f(y) then x= y" is true. Here f(x)= 2x^3+ 7x- 3 so we have 2x^3+ 7x- 3= 2y^3+ 7y- 3. Then 2(x^3- y^3)+ 7(x- y)= 2(x- y)(x^2+ xy+ y^2)+ 7(x- y)= (x- y)(2x^2+ 2xy+ 2y^2+ 7x- 7y)= 0. Either x- y= 0 or 2x^2+ 2xy+ 2y^2+ 7x- 7y)= 0. Use the quadratic formula to solve that second equation for x as a function of y. Show that there are no real x and y that satisfy that.