Graphing Exponents and logs.

Hi, I had two question assigned to me, both of which stumped me if some one would help it would be great.

Sketch a graph of the following and then fill in the table below:

f(x)=3x

g(x)=(3^2x)-3

h(x)=g-1(x)

Now i cant sketch them but for f and g i sketch them by imputing values for x and graph but for H I don't know how to graph that? it seems like a inverse function of G(x) but i cant quiet wrap my head around it.

The table asks for each function state

1) domain - All real numbers

2)range - All real numbers

3)asymptotes- i wrote no asymptotes for any

4)y intercept - 1)x=0 2)x=0.5 3) didn't get this far

5)x intercept - didn't get this far

Thanks in advance!

Re: Graphing Exponents and logs.

You can rewrite h(x) so that it doesn't contain g. $\displaystyle h(x)= (3^{2x})-3-x$

When sketching h(x) it is similar to your graph of g(x) but just subtracting x, redraw your sketch of g(x) but subtracting larger values of x as you move long the real line.

The y intercept occurs at x=0, your intercept for g(x) is incorrect.

The x intercept occurs at y=0, the x intercept might not exist, you can tell from your sketches.

Re: Graphing Exponents and logs.

My mistake, H(x) = (g^-1)(x) not g-1(x)

Re: Graphing Exponents and logs.

Anyone? how do i do this is its ^ the above comment

Re: Graphing Exponents and logs.

$\displaystyle g(x)=y=3^{2x}-3$

$\displaystyle y+3=3^{2x}$

$\displaystyle ln(y+3)=ln3^{2x}$

$\displaystyle ln(y+3)=2xln3$

$\displaystyle \frac{ln(y+3)}{2ln3}=x$

So then $\displaystyle g^{-1}(x)=\frac{ln(x+3)}{2ln3}$