Linearization of this equation / Inverse function

In summary, the conversation discusses finding the inverse function of a given equation with constants a, b, and c. The individual has tried using online tools but has not been successful, leading to the question of whether it is possible to find the inverse function. The summary then provides the solution for the inverse function, taking into account the constant 'a' multiplying the second term. The final result is f^-1(x)= (1/b) * LN((a(1-c)/(x-ac)).
  • #1
guiismiti
3
0
Hello,

I need to find the inverse function of the following equation

Code: 
y = a * ((exp(-b * x)) + (c * (1 - (exp(-b * x)))))

Where a, b and c are constants.

I have experimental points that fit to this equation and I want to use these values in the inverse funtion to linearize it.

I have tried to use a few tools available online, but the output functions did not work, which made me think if it is actually possible to do it.Can anybody help me?
Thanks in advance.

Edited: solved
Code: 
x = (1 / (-b)) * (LN(((a * c) - y) / (a * (c - 1))))
 
Last edited:
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  • #2
We are given:

\(\displaystyle f(x)=ae^{-bx}+c\left(1-e^{-bx}\right)=(a-c)e^{-bx}+c\)

To find the inverse function, we can write:

\(\displaystyle x=(a-c)e^{-by}+c\)

Solve for $y$:

\(\displaystyle x-c=(a-c)e^{-by}\)

\(\displaystyle \frac{x-c}{a-c}=e^{-by}\)

\(\displaystyle \ln\left(\frac{x-c}{a-c}\right)=-by\)

\(\displaystyle y=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)\)

Thus, we may claim:

\(\displaystyle f^{-1}(x)=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)\)
 
  • #3
MarkFL said:
We are given:

\(\displaystyle f(x)=ae^{-bx}+c\left(1-e^{-bx}\right)=(a-c)e^{-bx}+c\)

To find the inverse function, we can write:

\(\displaystyle x=(a-c)e^{-by}+c\)

Solve for $y$:

\(\displaystyle x-c=(a-c)e^{-by}\)

\(\displaystyle \frac{x-c}{a-c}=e^{-by}\)

\(\displaystyle \ln\left(\frac{x-c}{a-c}\right)=-by\)

\(\displaystyle y=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)\)

Thus, we may claim:

\(\displaystyle f^{-1}(x)=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)\)

The constant 'a' also multiplies the second term, that's why we got different results.

\(\displaystyle f(x)=ae^{-bx}+ac\left(1-e^{-bx}\right)\)
 
  • #4
guiismiti said:
The constant 'a' also multiplies the second term, that's why we got different results.

\(\displaystyle f(x)=ae^{-bx}+ac\left(1-e^{-bx}\right)\)

So it does...I missed that...lemme try again:

We are given:

\(\displaystyle f(x)=a\left(e^{-bx}+c\left(1-e^{-bx}\right)\right)=a(1-c)e^{-bx}+ac\)

To find the inverse function, we can write:

\(\displaystyle x=a(1-c)e^{-by}+ac\)

Solve for $y$:

\(\displaystyle x-ac=a(1-c)e^{-by}\)

\(\displaystyle \frac{x-ac}{a(1-c)}=e^{-by}\)

\(\displaystyle \ln\left(\frac{x-ac}{a(1-c)}\right)=-by\)

\(\displaystyle y=\frac{1}{b}\ln\left(\frac{a(1-c)}{x-ac}\right)\)

Thus, we may claim:

\(\displaystyle f^{-1}(x)=\frac{1}{b}\ln\left(\frac{a(1-c)}{x-ac}\right)\)
 
  • #5
MarkFL said:
So it does...I missed that...lemme try again

Done :)
 

FAQ: Linearization of this equation / Inverse function

What is linearization of an equation?

Linearization of an equation is a process of approximating a non-linear function with a linear function in a specific range. This is done by finding the tangent line to the curve at a given point, which serves as an approximation of the curve in that particular region.

Why is linearization important?

Linearization is important because it allows us to simplify complex non-linear functions and make them more manageable and easier to work with. It also helps in understanding the behavior of the function and making predictions about its values in a given range.

How do you linearize an equation?

To linearize an equation, you need to first choose a point on the curve and find its slope using the derivative. Then, using the point-slope form of a line, you can write the equation of the tangent line at that point. This equation represents the linearized form of the original equation.

What is an inverse function?

An inverse function is a function that undoes the effect of another function. In other words, if a function f(x) takes a number x and produces a result y, then the inverse function f-1(y) takes y and produces x. It is denoted as f-1(y) = x.

How do you find the inverse of a function?

To find the inverse of a function, you need to switch the positions of x and y in the original function and then solve for y. The resulting equation will be the inverse function. It is important to check if the inverse function is valid by making sure that the domain and range of the original function and its inverse are switched.

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