# Domain, Range, and Function Problems

• Jan 17th 2008, 01:47 PM
currypuff
Domain, Range, and Function Problems
1.) Please help me find the domain and range of $
h(x)=-\sqrt{x^2-36}
$

2.) For f(x) = 4x- $\sqrt{x}$ and g(x) = (3x+2)^2, please find f(g(x)).

• Jan 17th 2008, 01:56 PM
Plato
Did you graph these two functiona?
That will tell you a great deal.
• Jan 17th 2008, 06:25 PM
ThePerfectHacker
Quote:

Originally Posted by currypuff
1.) Please help me find the domain and range of $
h(x)=-\sqrt{x^2-36}
$

Domain: We must have $x^2 - 36\geq 0 \implies x\geq 6 \mbox{ or }x\leq -6$.

Range: It is all real $y$ such that the equation $y = - \sqrt{x^2 - 36}$ has a solution for $x$ in the domain. Thus, it means we want to solve, $-y = \sqrt{x^2 - 36}$ and RHS is always non-negative thus LHS must be non-negative and so we have a constraint that $-y \geq 0 \implies y\leq 0$. Square both sides, $y^2 = x^2 - 36 \implies x^2 = y^2 + 36$ since $y^2 + 36 \geq 0$ a square root exists and so the solution is $x = \pm \sqrt{y^2 + 36}$ but we want that $x\geq 6 \mbox{ or }x\leq -6$. If $x\geq 6$ then $\sqrt{y^2 + 36} \geq 6 \implies y^2 + 36 \geq 36 \implies y^2\geq 0$ which is always true. And similarly with $-\sqrt{y^2 + 36} \leq -6$. Thus, the range is $y\leq 0$.
• Jan 18th 2008, 09:29 AM
mathgeek777
Quote:

Originally Posted by currypuff
1.) Please help me find the domain and range of $
h(x)=-\sqrt{x^2-36}
$

This one is actually quite simple, you can set what is inside of the square root equal to zero and solve for x.

$-1(x^2-36) \geq0$

$-x^2 + 36 \geq 0$

$-x^2 \geq -36$

$x^2 \leq 36$

$\sqrt{x^2} \leq \sqrt{36}$

$x \leq 6$

There's your domain. Not sure how to find the range, but I'm sure someone here can help you on that :)

EDIT: Dag nabbit! Hacker beat me to it again!!!
• Jan 18th 2008, 09:46 AM
topsquark
Quote:

Originally Posted by mathgeek777
$x \leq 6$

There's your domain. Not sure how to find the range, but I'm sure someone here can help you on that :)

That would be
$-6 \leq x \leq 6$
actually. ;)

-Dan