# Inverse and differentation of an equation

• MHB
• ertagon2
In summary: 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 $4$.
ertagon2
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How do I solve this. Getting an inverse alone seems to be a quite long route. Is there an easier way of doing this?

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ertagon2 said:
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. :)

I would use:

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

Can you proceed?

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.)

MarkFL said:
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. :)

I would use:

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

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)?

ertagon2 said:
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 $b$?

MarkFL said:
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?

ertagon2 said:
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. :)

Krylov said:
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?

MarkFL said:
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.

MarkFL said:
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.

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?

ertagon2 said:
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.

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ertagon2 said:
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?

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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?

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

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

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:

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

Next, we solve:

$$\displaystyle f(x)=-3$$

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

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

The only real root is:

$$\displaystyle x=0$$

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

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

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.

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## 1. What is the inverse of an equation?

The inverse of an equation is a function that "undoes" the original function. In other words, if f(x) is the original function, then its inverse, denoted as f-1(x), will return the input value that results in a given output value. This is also known as "reversing" the function.

## 2. How do you find the inverse of an equation?

To find the inverse of an equation, you can follow these steps:
1. Write the original equation in the form y = f(x).
2. Switch the x and y variables.
3. Solve for y to get the inverse function, f-1(x).
Note: The resulting inverse function might not be a function if the original function is not one-to-one (i.e. each input has only one corresponding output).

## 3. What is the relationship between the inverse and derivative of an equation?

The inverse and derivative of an equation are related in that the derivative of the inverse function is the reciprocal of the derivative of the original function. In other words, if f(x) and f-1(x) are inverse functions, then (f-1)'(x) = 1/f'(x). This is known as the Inverse Function Theorem.

## 4. How do you differentiate an inverse function?

To differentiate an inverse function, you can use the Inverse Function Theorem. If y = f(x) and x = f-1(y), then by differentiating both equations with respect to x, we get:
1 = f'(f-1(y)) * (f-1)'(y)
Solving for (f-1)'(y), we get:
(f-1)'(y) = 1/f'(f-1(y))
So, to differentiate an inverse function, we simply take the reciprocal of the derivative of the original function evaluated at the inverse function's input value.

## 5. Why is it important to understand inverse and differentiation of equations?

Understanding inverse and differentiation of equations is important in many fields of science and mathematics. Inverse functions are used to "undo" certain operations and can help in solving equations, finding roots, and graphing functions. Differentiation, on the other hand, is a fundamental concept in calculus and is used to find rates of change, optimize functions, and solve many problems in physics, engineering, and economics. By understanding these concepts, we can better analyze and model real-world situations and make predictions based on data.

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