Suppose that the functions g and h are defined as follows. g(x)=-5+4x^2 h(x)=4-9x (a) Find ((g)/(h))(3) (b) Find all values that are NOT in the domain of (g)/(h) If there is more than one value, separate them with commas. (a) ((g)/(h))(3)=square (b) Value(s) that are NOT in the domain of (g)/(h):square

(a) To find $(\frac {g}{h})(3)$, we need to substitute $x=3$ into both $g(x)$ and $h(x)$ and then divide the result of $g(x)$ by the result of $h(x)$.<br /><br />First, let's find $g(3)$:<br />$g(3) = -5 + 4(3)^2 = -5 + 4(9) = -5 + 36 = 31$<br /><br />Next, let's find $h(3)$:<br />$h(3) = 4 - 9(3) = 4 - 27 = -23$<br /><br />Now, we can find $(\frac {g}{h})(3)$:<br />$(\frac {g}{h})(3) = \frac{g(3)}{h(3)} = \frac{31}{-23} = -\frac{31}{23}$<br /><br />Therefore, $(\frac {g}{h})(3) = -\frac{31}{23}$.<br /><br />(b) The values that are NOT in the domain of $\frac {g}{h}$ are the values for which $h(x) = 0$, since division by zero is undefined.<br /><br />To find these values, we set $h(x) = 0$ and solve for $x$4 - 9x = 0$<br />$9x = 4$<br />$x = \frac{4}{9}$<br /><br />Therefore, the value that is NOT in the domain of $\frac {g}{h}$ is $\frac{4}{9}$.<br /><br />In summary:<br />(a) $(\frac {g}{h})(3) = -\frac{31}{23}$<br />(b) Value(s) that are NOT in the domain of $\frac {g}{h}$: $\frac{4}{9}$