1.) Please help me find the domain and range of $\displaystyle
h(x)=-\sqrt{x^2-36}
$
2.) For f(x) = 4x-$\displaystyle \sqrt{x}$ and g(x) = (3x+2)^2, please find f(g(x)).
Thanks in advance for your assistance.
Domain: We must have $\displaystyle x^2 - 36\geq 0 \implies x\geq 6 \mbox{ or }x\leq -6$.
Range: It is all real $\displaystyle y$ such that the equation $\displaystyle y = - \sqrt{x^2 - 36}$ has a solution for $\displaystyle x$ in the domain. Thus, it means we want to solve, $\displaystyle -y = \sqrt{x^2 - 36}$ and RHS is always non-negative thus LHS must be non-negative and so we have a constraint that $\displaystyle -y \geq 0 \implies y\leq 0$. Square both sides, $\displaystyle y^2 = x^2 - 36 \implies x^2 = y^2 + 36$ since $\displaystyle y^2 + 36 \geq 0$ a square root exists and so the solution is $\displaystyle x = \pm \sqrt{y^2 + 36}$ but we want that $\displaystyle x\geq 6 \mbox{ or }x\leq -6$. If $\displaystyle x\geq 6$ then $\displaystyle \sqrt{y^2 + 36} \geq 6 \implies y^2 + 36 \geq 36 \implies y^2\geq 0$ which is always true. And similarly with $\displaystyle -\sqrt{y^2 + 36} \leq -6$. Thus, the range is $\displaystyle y\leq 0$.
This one is actually quite simple, you can set what is inside of the square root equal to zero and solve for x.
$\displaystyle -1(x^2-36) \geq0$
$\displaystyle -x^2 + 36 \geq 0$
$\displaystyle -x^2 \geq -36$
$\displaystyle x^2 \leq 36$
$\displaystyle \sqrt{x^2} \leq \sqrt{36}$
$\displaystyle x \leq 6$
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!!!