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!!
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