How can you use a point on the graph of f –1(x) = 9x to determine a point on the graph of f(x) = log9x?

Whenever you can invert a function, you have that the graphs of and its inverse are reflected with respect to the line

So, given any point on the graph of , you can simply swap its coordinates to get the correspondent point on the original function

As an example, all exponential functions pass through the point , while all the logarithmic functions pass through the point

Switch the x- and y-coordinates

Step-by-step explanation:

Answer A on Edge-nuity.

step-by-step explanation: step 3 should really go to step seven, but anyway never mind.

1 ∠oln ≅ ∠lno