Finding the inverse of function

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Homework Statement



find the inverse, f(x) = 1 + e^sinh(x)

Homework Equations


The Attempt at a Solution


I am sorry but I never encounter this problem before and my teacher never showed us how to do these kind of problems, I have no idea what to do or how to start it out =[
 
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start with the equation of sinh(x)

then let u = e^{x} and see if you can solve for u
 
so is it: (u - e^-x) / 2 = 0 ?
then: u = e^-x ?
 
yojo95 said:

Homework Statement



find the inverse, f(x) = 1 + e^sinh(x)

Homework Equations





The Attempt at a Solution


I am sorry but I never encounter this problem before and my teacher never showed us how to do these kind of problems, I have no idea what to do or how to start it out =[
I seriously doubt that your teacher "never showed us how to do these kind of problems". Perhaps not with these particular functions- but the idea is the same as for linear functions. To find the inverse of any function f(x), write y= f(x), then "swap" x and y, x= f(y), and solve for y. If you have ever learned how to solve equations, you have learned how to do this.

(To shortcut any arguments, yes, some people learn to "first solve y= f(x) for x, then swap x and y. It's the same thing.)

y= f(x)= 1+ e^{sinh(x)}
becomes
x= 1+ e^{sinh(y)}

Solve for y by "backing out". Since 1 is added on the right, subtract 1 from each side:
x- 1= e^{sinh(y)}
Now we have an exponential on the right. The opposite of that is "ln" so take the natural logarithm of both sides:
ln(x-1)= sinh(y)[/itex]<br /> <br /> What do you think we should do now?
 
since we want to get rid of sinh, do we take the inverse sinh, sinh-1, of both sides?

sinh-1(ln(x-1)) = sinh-1(sinh(y))
sinh-1(ln(x-1)) = y
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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