a.) find the domain and range of g(x)

b.) find g^-1(x)

c.) find the domain and range of g^-1(x)

a) the square root can never be negative

So, 3 - 5x >/= 0

-5x >/= -3

x </= 3/5 that's the domain

b) to find g^-1(x), we switch the x and y variables

x = sqrt(3 - 5y)

Now we solve for y:

x^2 = 3 - 5y

5y = 3 - x^2

y = 3/5 - x^2/5 or -1/5x^2 + 3/5

Therefore, g^-1(x) = -1/5x^2 + 3/5

c) the domain of g^-1(x) is ALL REAL NUMBERS

since x^2 is always positive, the smallest it can be is zero

therefore, 0 + 3/5 = 3/5 is the highest the range can be. It goes down from there as the -1/5x^2 term becomes more negative

therefore, the range is -infinity < y </= 3/5

Thank you again. I am studying for a test tomorrow and want to make sure that I have studied for everything. Again thank you for your help!

You're most welcome :) Good luck tomorrow!!

Thanks again!

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