How do the graphs of the functions f(x) = (3/2)^x and g(x) = (2/3)^2 compare?

First of all, you have a small typo in your question. The actual question is: How do the graphs of the functions f(x) = (3/2)^x and g(x) = (2/3)^x compare?

g(x) = (2/3)^2 is not even a function :)

Notice that the base of g(x) is the base of f(x) inverted. Every time you see that pattern expect a reflection about the y-axis.

Let's check it out why.

There are two types of reflections in exponential function

- Reflection about the x-axis

To reflect an exponential function about the x-axis, multiply the output of the function by -1. For example, to reflect the function f(x)=e^x about the x-axis, we multiply the output, e^2, of function by -1 to get f(x)=-e^x.

- Reflection about the y-axis.

To reflect the exponential function about the y-axis, multiply the input of the function by -1. For example, to reflect the function f(x)=e^x, we multiply the input, x, of the function by -1 to get f(x)=e^-x.

Let's multiply the output our of function f(x) and see what happens:

Remember that according to the laws of exponents to get rid of the minus sign in the exponent we just need to flip the fraction:

Look! We just transformed f(x) into g(x)! Proving that they are reflections of each other over the y-axis.

We can conclude that the correct answer is the graphs are reflections of each other over the y-axis.

Here, In this question, your options are:
The graphs are reflections of each other over the y-axis. 
The g(x) function represents exponential decay. 
The f(x) function represents exponential growth. 
They have the same initial value.

Here, You can see in f(x) = (3/2)^x, where exponential is variable, which can be anything either positive or negative.

In function g(x) = (2/3)^2, denominator is greater than numerator, so their square will be smaller than the number itself. Hence, 2 is in power, it will show exponential decay.

In short, Your Answer would be Option B

Hope this helps!

Ignore the question it is wrong

Step-by-step explanation: