domain and 1 to 1..ps dan you rock

• Mar 30th 2006, 11:05 AM
mazdatuner
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)
• Mar 31st 2006, 04:13 AM
topsquark
Quote:

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! :)

$\displaystyle F(x)=\sqrt{x-3}$ and $\displaystyle G(x)=4x^2+3$

a) You need to find: $\displaystyle (G \circ F)(x)=G(F(x))$
So.
$\displaystyle G(F(x)) = G(\sqrt{x-3})=4(\sqrt{x-3})^2+3$
$\displaystyle =4(x-3)+3=4x-9$
(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 $\displaystyle (-\infty,\infty)$ because there are no (apparent) restrictions on the domain. However, there IS a restriction given by the F(x): $\displaystyle x \ge 3$. 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 $\displaystyle 4x-9$ and $\displaystyle 4(\sqrt{x-3})^2+3$, which of course there is.

-Dan
• Mar 31st 2006, 04:21 AM
topsquark
Quote:

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 $\displaystyle y=x^2+4$. It's a tad more convenient. To get the inverse function simply switch the position of x and y, then solve for y.

$\displaystyle x=y^2+4$

$\displaystyle y^2=x-4$

$\displaystyle y=\sqrt{x-4}$
Technically we should have a $\displaystyle \pm$ 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 $\displaystyle G^{-1}(x)=\sqrt{x-4}$.

(A good way to check this is to calculate $\displaystyle G(G^{-1}(x))$ and $\displaystyle G^{-1}(G(x))$, 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 $\displaystyle [4, \infty)$.

The range is all the values of the function over its domain, so I get $\displaystyle [0, \infty)$.

-Dan
• Mar 31st 2006, 08:49 AM
ThePerfectHacker
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 $\displaystyle G^{-1}=\sqrt{x-4}$ 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:
$\displaystyle y=\sqrt{x-4}$
has a real solution for x.
Notice that $\displaystyle \sqrt{x-4}\geq 0$ thus, there would be no solutions if $\displaystyle y<0$. Now, when $\displaystyle y\geq 0$ then you can square both sides to get.
$\displaystyle y^2=x-4$ and notice this equation has a solution $\displaystyle x=y^2+4$. Thus, all and only those values of $\displaystyle y$ that are non-negative are in the range. Thus, $\displaystyle y\geq 0$ or another way of writing this is $\displaystyle y\in [0,\infty]$