find the derivative of the function using the definition of derivative. ( g(t) = \frac{1 - 2t}{6 + t} ) ( g(t) = ) blank state the domain of the function. (enter your answer using interval notation.) blank state the domain of its derivative. (enter your answer using interval notation.) blank

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Accepted Answer

### Part 1: Find the derivative $ G'(t) $ using the definition of derivative

The definition of the derivative of a function $ G(t) $ is:
$$
G'(t) = \lim_{h \to 0} \frac{G(t + h) - G(t)}{h}
$$

#### Step 1: Compute $ G(t + h) $
Given $ G(t) = \frac{1 - 2t}{6 + t} $, we substitute $ t + h $ into the function:
$$
G(t + h) = \frac{1 - 2(t + h)}{6 + (t + h)} = \frac{1 - 2t - 2h}{6 + t + h}
$$

#### Step 2: Compute $ G(t + h) - G(t) $
$$
\begin{align*}
G(t + h) - G(t) &= \frac{1 - 2t - 2h}{6 + t + h} - \frac{1 - 2t}{6 + t} \
&= \frac{(1 - 2t - 2h)(6 + t) - (1 - 2t)(6 + t + h)}{(6 + t + h)(6 + t)}
\end{align*}
$$

Expand the numerator:
$$
\begin{align*}
&(1 - 2t - 2h)(6 + t) - (1 - 2t)(6 + t + h) \
&= (1)(6 + t) - 2t(6 + t) - 2h(6 + t) - [ (1)(6 + t + h) - 2t(6 + t + h) ] \
&= 6 + t - 12t - 2t^2 - 12h - 2ht - [6 + t + h - 12t - 2t^2 - 2ht] \
&= 6 - 11t - 2t^2 - 12h - 2ht - 6 - t - h + 12t + 2t^2 + 2ht \
&= (6 - 6) + (-11t - t + 12t) + (-2t^2 + 2t^2) + (-12h - h) + (-2ht + 2ht) \
&= -13h
\end{align*}
$$

So, $ G(t + h) - G(t) = \frac{-13h}{(6 + t + h)(6 + t)} $

#### Step 3: Divide by $ h $ and take the limit as $ h \to 0 $
$$
\begin{align*}
\frac{G(t + h) - G(t)}{h} &= \frac{\frac{-13h}{(6 + t + h)(6 + t)}}{h} \
&= \frac{-13}{(6 + t + h)(6 + t)}
\end{align*}
$$

Now, take the limit as $ h \to 0 $:
$$
G'(t) = \lim_{h \to 0} \frac{-13}{(6 + t + h)(6 + t)} = \frac{-13}{(6 + t)(6 + t)} = \frac{-13}{(6 + t)^2}
$$

### Part 2: State the domain of the function $ G(t) $
The function $ G(t) = \frac{1 - 2t}{6 + t} $ is a rational function. The denominator cannot be zero:
$$
6 + t
eq 0 \implies t
eq -6
$$
So, the domain of $ G(t) $ is $ (-\infty, -6) \cup (-6, \infty) $

### Part 3: State the domain of its derivative $ G'(t) $
The derivative $ G'(t) = \frac{-13}{(6 + t)^2} $ is also a rational function. The denominator $ (6 + t)^2 $ cannot be zero:
$$
(6 + t)^2
eq 0 \implies 6 + t
eq 0 \implies t
eq -6
$$
So, the domain of $ G'(t) $ is $ (-\infty, -6) \cup (-6, \infty) $

### Final Answers

1. Derivative: $ \boldsymbol{G'(t) = \frac{-13}{(6 + t)^2}} $
2. Domain of $ G(t) $: $ \boldsymbol{(-\infty, -6) \cup (-6, \infty)} $
3. Domain of $ G'(t) $: $ \boldsymbol{(-\infty, -6) \cup (-6, \infty)} $

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QUESTION IMAGE

Question

Explanation:

Part 1: Find the derivative \( G'(t) \) using the definition of derivative

Step 1: Compute \( G(t + h) \)

Step 2: Compute \( G(t + h) - G(t) \)

Step 3: Divide by \( h \) and take the limit as \( h \to 0 \)

Part 2: State the domain of the function \( G(t) \)

Part 3: State the domain of its derivative \( G'(t) \)

Final Answers

Answer: