Let #h(t) = (2-t)/(t)# and #g(t)= (3t+15)/t#, how do you find (h/g)(t) and the domain for h/g?

Question

Caleb Alexander · Accepted Answer

We have that #(h(t))/g(t)=(2-t)/(3t+15)# and the domain of #(h(t))/g(t)# is #R-{0,-5}#

We have that

#(h(t))/g(t)=((2-t)/t)/((3t+15)/t)=(2-t)/(3t+15)#

The domain of #(h(t))/g(t)# is #R-{0,-5}#

The domain of a quotient is a subset of the intersection of the domains. Since neither domain contains 0, the domain of the quotient excludes 0 as well.

Ariana Alvarez · Answer

undefined To find $ \frac{h(t)}{g(t)} $ and determine its domain, follow these steps:

1. **Write the Expression for $ \frac{h(t)}{g(t)} $**:
$$ \frac{h(t)}{g(t)} = \frac{\frac{2-t}{t}}{\frac{3t+15}{t}} $$

2. **Simplify the Expression**:
$$ \frac{h(t)}{g(t)} = \frac{2-t}{3t+15} $$

3. **Determine the Domain**:
The domain is determined by the values of $ t $ for which the expression is defined. In this case, since we have a rational expression, we need to exclude values of $ t $ that make the denominator zero.
$$ 3t + 15 = 0 $$
$$ 3t = -15 $$
$$ t = -5 $$

Thus, the domain for $ \frac{h(t)}{g(t)} $ is all real numbers except $ t = -5 $, or in interval notation: $ (-\infty, -5) \cup (-5, \infty) $.

Benjamin Achtenberg · Answer

undefined To find $(h/g)(t)$, we divide $h(t)$ by $g(t)$ term-wise.

$$
\frac{h(t)}{g(t)} = \frac{\frac{2-t}{t}}{\frac{3t+15}{t}} = \frac{2-t}{3t+15}
$$

The domain for $(h/g)(t)$ will be the intersection of the domains of $h(t)$ and $g(t)$, excluding any values of $t$ that make the denominator of $g(t)$ equal to zero. Since the denominator of $g(t)$ is $t$, we cannot have $t = 0$. Therefore, the domain for $(h/g)(t)$ is all real numbers except $t = 0$.