Question

Calculus 2...

Answer 1

Consider the following.\[f(x)=\frac{ a^x-1 }{ a^x+1 } \]
\[a>0, a \neq1\]
Show that f has an inverse function. Then find f^-1

f^-1(x)=????

Answer 2

You can think of \(f(x)\) as \(y\). Then you just swap \(x\) and \(y\) and solve for \(y\).
\[ \frac{a^y-1}{a^y+1}=x\\
\,\\
\frac{a^y-1}{x}=a^y+1\\
\,\\
\frac{a^y}{x}-\frac{1}{x}=a^y+1\\
\, \\
\frac{a^y}{x}-\frac{1}{x}-a^y=1\\
\, \\
a^y\left(\frac{1}{x}-1 \right)=\frac{1}{x}+1\\
\, \\
\log \left[a^y\left(\frac{1}{x}-1 \right)\right]=\log\left(\frac{1}{x}+1 \right)\\
\, \\
\log a^y+\log\left( \frac{1}{x}-1\right)=\log\left( \frac{1}{x}+1\right)\\
\, \\
y\log a+\log\left( \frac{1-x}{x}\right)=\log\left(\frac{1+x}{x}\right)\\
\, \\
y\log a =\log\left(\frac{1+x}{x}\right)-\log\left( \frac{1-x}{x}\right)\\
y\log a=\log\left( \frac{\frac{1+x}{x}}{\frac{1-x}{x}}\right)\\
\, \\
y=\frac{\log\left( \frac{1+x}{1-x}\right)}{\log a}\]

Answer 3

Thank you so much! Most people here don't explain it clearly like you did, they just tell you to do this and that and I have no idea what they are talking about

Answer 4

My only question is why do need to bring up log for?

Answer 5

:)

Answer 6

Oh because you need to solve for y! You have an exponential form \(a^y\) and you need to solve for the y. If you think of a simple exponential like
\(3^x=7\), how do you solve for \(x\)? You log both sides and use the properties of logs.

Answer 7

You just blew my mind...Calculus makes so much sense now (not exaggerating)! THANK YOU SO MUCH!

Answer 8

I've always wondered how to solve something like that

Answer 9

oh haha I am glad then :)