As you approach the seventh gate, you notice that it is only a short distance to the area where your computer
indicated the communication device was stored. You look past the gate hopefully, as your computer translates:
The passcode for this gate is found by determining the domain of the derivative of and then evaluating
the derivative at . Enter the domain in interval notation.
.
Domain of :

Ask by Huang O'Quinn. in the United States

Mar 30,2025

Question

As you approach the seventh gate, you notice that it is only a short distance to the area where your computer
indicated the communication device was stored. You look past the gate hopefully, as your computer translates:
The passcode for this gate is found by determining the domain of the derivative of and then evaluating
the derivative at . Enter the domain in interval notation.
.
Domain of :

Ask by Huang O'Quinn. in the United States

Mar 30,2025

UpStudy AI · Accepted Answer

The given function is $$ f(x) = \sqrt{4 - 82x} $$. To find the domain of the derivative, we first need to determine the domain of $$ f(x) $$ itself.
For the square root function $$ f(x) $$ to be defined, the expression inside the square root must be non-negative:

$$
4 - 82x \geq 0
$$

Solving this inequality will give us the domain of $$ f(x) $$.
Rearranging the inequality:

$$
4 \geq 82x
$$

Dividing both sides by 82 (noting that 82 is positive, so the inequality direction remains the same):

$$
\frac{4}{82} \geq x
$$

This simplifies to:

$$
x \leq \frac{2}{41}
$$

Thus, the domain of $$ f(x) $$ in interval notation is:

$$
(-\infty, \frac{2}{41}]
$$
Next, we will find the derivative $$ f'(x) $$ using the chain rule. The derivative of $$ f(x) = \sqrt{g(x)} $$ where $$ g(x) = 4 - 82x $$ is given by:

$$
f'(x) = \frac{1}{2\sqrt{g(x)}} \cdot g'(x)
$$

Calculating $$ g'(x) $$:

$$
g'(x) = -82
$$

Thus, the derivative is:

$$
f'(x) = \frac{-82}{2\sqrt{4 - 82x}} = \frac{-41}{\sqrt{4 - 82x}}
$$
For $$ f'(x) $$ to be defined, the expression under the square root must be positive:

$$
4 - 82x > 0
$$

This leads to the same inequality as before, but we need strict inequality:

$$
x < \frac{2}{41}
$$

Thus, the domain of $$ f'(x) $$ in interval notation is:

$$
(-\infty, \frac{2}{41})
$$
Now we will evaluate the derivative at $$ x = -\frac{60}{82} $$:

First, we simplify $$ -\frac{60}{82} $$:

$$
-\frac{60}{82} = -\frac{30}{41}
$$

Since $$ -\frac{30}{41} < \frac{2}{41} $$, it is within the domain of $$ f'(x) $$. Now we substitute $$ x = -\frac{30}{41} $$ into the derivative:

$$
f'\left(-\frac{30}{41}\right) = \frac{-41}{\sqrt{4 - 82\left(-\frac{30}{41}\right)}}
$$

Calculating the expression inside the square root:

$$
4 - 82\left(-\frac{30}{41}\right) = 4 + \frac{2460}{41} = \frac{164 + 2460}{41} = \frac{2624}{41}
$$

Now substituting back into the derivative:

$$
f'\left(-\frac{30}{41}\right) = \frac{-41}{\sqrt{\frac{2624}{41}}} = \frac{-41 \cdot \sqrt{41}}{\sqrt{2624}}
$$
The domain of $$ f'(x) $$ is $$ (-\infty, \frac{2}{41}) $$. The evaluation of $$ f'\left(-\frac{60}{82}\right) $$ results in $$ \frac{-41 \cdot \sqrt{41}}{\sqrt{2624}} $$.

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