1. ## inverse function question..

i need to find the inverse function of these:

ImageShack - Image Hosting :: 63348338nf8.gif

i know that if y=sinx
then its inverse whould be x=arcsin y

but the first case has two = signs
and the second one is a split function

what to do in these cases?

2. Originally Posted by transgalactic
i need to find the inverse function of these:

ImageShack - Image Hosting :: 63348338nf8.gif

i know that if y=sinx
then its inverse whould be x=arcsin y

but the first case has two = signs
and the second one is a split function

what to do in these cases?
In the first one you just have to find the inverse of tanh(x), given that tanh(x) = all those e things. Given that natural logarithm (ln(x)) is the inverse of (e^x), I suspect that the inverse of tanh will be a function of ln(x).

In the 2nd piecewise function you just have to find the inverse of each piece as if it were a separate function.

For example f(x) = x^2 1 < x < 4 would lead to the inverse, g(x) being:

g(x) = sqrt(x) 1<x<4

3. regarding the first "hiperbolic" question

there are two "=" signs
thats not a equation

i dont know how to get one equation with one "=" sign out of it?

4. Originally Posted by transgalactic
regarding the first "hiperbolic" question

there are two "=" signs
thats not a equation

i dont know how to get one equation with one "=" sign out of it?
Lol there's nothing funny going on, I promise you!

All it's saying it this:

$f(x) = tanh(x)$

AND

$tanh(x) = \frac{e^x - e^{-x}}{e^x+e^{-x}}$

THEREFORE:

$f(x) = \frac{e^x - e^{-x}}{e^x+e^{-x}}$

What is $f^{-1}(x)$ ?

An "equation" means, two eqate to things. Or to state them to be equal. Therefore, an equation can have as many equal signs as it wants. As long as what it's saying is TRUE, then it's still an equation. This is an equation:

$5 = 4+1=3+2=2+2+1=1+1+1+2=3+1+1$

5. Originally Posted by Mush
$f(x) = \frac{e^x - e^{-x}}{e^x+e^{-x}}$

What is f^{-1}(x) ?
$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-\frac{1}{e^x}}{e^x+\frac{1}{e^x}}$. Let $z=e^x$ and this becomes $y=\frac{z-\frac{1}{z}}{z+\frac{1}{z}}$ multiplying both sides by $z+\frac{1}{z}$ gives $y\left(z+\tfrac{1}{z}\right)=z-\tfrac{1}{z}$. Multiplying both sides by $z$ gives $y(z^2+1)=(z^2-1)$ or equivalently $yz^2+y-z^2+1=0$. Factoring gives $(y-1)z^2+(y+1)=0$. Solving gives $z=\pm\sqrt{\frac{y+1}{y-1}}$. Which implies that $e^x=\pm\sqrt{\frac{y+1}{y-1}}$ or $x=\ln\left(\pm\sqrt{\frac{y+1}{y-1}}\right)=\text{arctanh}(y)$

6. Originally Posted by Mathstud28
$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-\frac{1}{e^x}}{e^x+\frac{1}{e^x}}$. Let $z=e^x$ and this becomes $y=\frac{z-\frac{1}{z}}{z+\frac{1}{z}}$ multiplying both sides by $z+\frac{1}{z}$ gives $y\left(z+\tfrac{1}{z}\right)=z-\tfrac{1}{z}$. Multiplying both sides by $z$ gives $y(z^2+1)=(z^2-1)$ or equivalently $yz^2+y-z^2+1=0$. Factoring gives $(y-1)z^2+(y+1)=0$.

Solving gives $z=\pm\sqrt{\frac{y+1}{y-1}}$. Mr F says: No. ${\color{red}z=\pm\sqrt{\frac{y+1}{{\color{blue}1-y}}}}$. But z > 0 since ${\color{red}z = e^x}$ and x is real. Therefore ${\color{red}z = \sqrt{\frac{y+1}{1-y}}}$.

[snip]
There some errors (see red). Once these corrections are made, the solution will continue in a similar way to that already shown.