a. …evaluate and simplify h(0) h(x) = -15-3x ______ 12-6x b. …what is the domain of h ? (Be sure to use set notation since domains are sets!)
Follow Math Help Forum on Facebook and Google+
Originally Posted by mjfaris a. …evaluate and simplify h(0) h(x) = -15-3x ______ 12-6x b. …what is the domain of h ? (Be sure to use set notation since domains are sets!) $\displaystyle h(0)=\frac{-15-3(0)}{12-6(0)}= \frac{-15}{12} = \frac{-5}{4}$ b) the deminator should not equal 0 so $\displaystyle 12-6x = 0 \Rightarrow x=2 $ so x all real numbers except 0 as an interval $\displaystyle (-\infty, 0)\cup (0,\infty)$
Originally Posted by Amer $\displaystyle h(0)=\frac{-15-3(0)}{12-6(0)}= \frac{-15}{12} = \frac{-5}{4}$ b) the deminator should not equal 0 so $\displaystyle 12-6x = 0 \Rightarrow x=2 $ so x all real numbers except 0 as an interval $\displaystyle (-\infty, 0)\cup (0,\infty)$ You mean, surely, "so x all real numbers except 2. As an interval $\displaystyle (-\infty, 2)\cup (2, \infty)$
View Tag Cloud