Thread: domain and 1 to 1..ps dan you rock

id like to thank you dan for your previous help.....it really helped me understand how the process worked

I have 2 questions if anyone could help me

with F(x) = square root of x - 3 and G(x) = 4x^2 + 3
a) write the function (G of F)(X) thats G of F of X
I came up with G(square root of x - 3) = 4(square root of x - 3)^2 + 3
G(x) = 4x + 12
b) find the domain...of the G of F of X ....
i have a few ideas but am kinda lost on this one




SECOND QUESTION
Given the one to one function G(x)= x^2+4 where X >
write g^-1(x)
write domain and range of G(x) G^-1(x)

Originally Posted by mazdatuner

F(x) = square root of x - 3 and G(x) = 4x^2 + 3
a) write the function (G of F)(X) thats G of F of X
I came up with G(square root of x - 3) = 4(square root of x - 3)^2 + 3
G(x) = 4x + 12
b) find the domain...of the G of F of X ....
i have a few ideas but am kinda lost on this one

You are very welcome!

and

a) You need to find:
So.


(You forgot to add the last 3.)

b) You need to be a bit careful when considering the domain. At first blush it looks like the domain is because there are no (apparent) restrictions on the domain. However, there IS a restriction given by the F(x): . So we can't actually have any x less than 3 in G(F(x)). This is subtle and implies that there actually IS a distinction between and , which of course there is.

-Dan

Originally Posted by mazdatuner

Given the one to one function G(x)= x^2+4 where X >
write g^-1(x)
write domain and range of G(x) G^-1(x)

I usually like to skip the G(x) notation here and set . It's a tad more convenient. To get the inverse function simply switch the position of x and y, then solve for y.






Technically we should have a in front of this, but we have to choose one or the other, else we no longer have a function. The choice of + is standard. So .

(A good way to check this is to calculate and , which both should be equal to x.)

The domain of this is x such that the expression is real (and not otherwise undefined.) So the domain is .

The range is all the values of the function over its domain, so I get .

-Dan

I realized many people have difficulty finding the range of the function. The usual method is to sketch the function and see what values y-takes. But, the problem is it non-rigorous. Here is an algebraic method for which you can find the range. Since you found what you need to find its range. By definition a certain value of y is in the range of the function if and only if there is such an x when mapped by the function gives y. Thus, you need to find all y such the the equation:

has a real solution for x.
Notice that thus, there would be no solutions if . Now, when then you can square both sides to get.
and notice this equation has a solution . Thus, all and only those values of that are non-negative are in the range. Thus, or another way of writing this is