Just a little quickie, since Google won't help me out.

g(x) = 9^(x)-3

h(x) = g^(-1)(x)

Is g^(-1)(x) equal to (1/g)(x) or 1/(g(x))? In other words, would g^(-1) be equivalent to 1 over all of 9^(x)-3?

Aleguan Jan 24, 2018

#1**+3 **

In this context g^{-1}(x) does not mean g(x) raised to the power of negative 1 .

It means the inverse function of g(x) . We can find g^{-1}(x) like this....

g(x) = 9^x - 3

Instead of g(x) , write y .

y = 9^x - 3

Now we want to solve this equation for x .

y + 3 = 9^x

log(y + 3) = log(9^x)

log(y + 3) = x log 9

log(y + 3) / log 9 = x

To get the inverse function, swap x and y .

log(x + 3) / log 9 = y This is the inverse function. We can say...

g^{-1}(x) = log(x + 3) / log 9

So... g^{-1}(x) does not mean 1 / g(x) .

Neither does sin^{-1}x = (sin x)^{-1} , even though sin^{2}x = (sin x)^{2}

This is something that has bothered me before. Maybe there is a reason why it is okay and I just don't understand it......?

hectictar Jan 24, 2018

#3**+2 **

"log" is short for "logarithm" .

This can probably explain better than I can: https://youtu.be/Z5myJ8dg_rM

Here's a simpler example of finding the inverse function:

f(x) = x + 1

Instead of f(x) , let's say y .

y = x + 1 Solve for x .

y - 1 = x

Swap x and y .

x - 1 = y This is the inverse.

f^{-1}(x) = x - 1

hectictar Jan 24, 2018