Let's go through both problems shown in the image:

Problem 36: Drug Dosages

We are given the formula for pediatric drug dosages using Cowling's rule:

$D(t) = \left( \frac{t + 1}{24} \right) a$

where:

• $t$ is the child's age (in years),
• $a$ is the adult dosage (in milligrams),
• $D(t)$ is the child's dosage.

Part a: Show that $D(t)$ is a linear function of $t$.

The formula provided is:

$D(t) = \left( \frac{t + 1}{24} \right) a$

We can rewrite it as:

$D(t) = \frac{t}{24}a + \frac{a}{24}$

This is in the form of a linear function:

$D(t) = mt + b$

where:

• $m = \frac{a}{24}$ (this is the slope),
• $b = \frac{a}{24}$ (this is the y-intercept).

Therefore, $D(t)$ is indeed a linear function of $t$.

Part b: If the adult dose of a drug is 500 mg, how much should a 4-year-old child receive?

Using the formula:

$D(t) = \left( \frac{t + 1}{24} \right) a$

Substitute $t = 4$ and $a = 500$:

$D(4) = \left( \frac{4 + 1}{24} \right) \times 500 = \left( \frac{5}{24} \right) \times 500 = \frac{2500}{24} \approx 104.17 \, \text{mg}$

So, a 4-year-old child should receive approximately 104.17 mg of the drug.

Problem 4: Analyze the function $f(x) = x^2 - x - 6$

We need to find:

1. Intervals of increase and decrease: These are determined by finding the derivative and checking where it is positive (increasing) and negative (decreasing).
2. Critical points and turning points: Where the derivative equals zero.
3. Intercepts: Both x- and y-intercepts.
4. Maximum and minimum values.

Let's analyze each step:

1. Derivative of the function: $f'(x) = 2x - 1$ Setting this equal to zero gives: $2x - 1 = 0 \implies x = \frac{1}{2}$ So, $x = \frac{1}{2}$ is the critical point.

2. Intervals of increase/decrease:

   • For $x < \frac{1}{2}$, $f'(x) < 0$, so the function is decreasing.
   • For $x > \frac{1}{2}$, $f'(x) > 0$, so the function is increasing.

3. Intercepts:

   • y-intercept: $f(0) = -6$, so the y-intercept is $(0, -6)$.
   • x-intercepts: Solve $x^2 - x - 6 = 0$ using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm \sqrt{25}}{2}$ $x = \frac{1 \pm 5}{2}$ So, $x = 3$ and $x = -2$. The x-intercepts are $(-2, 0)$ and $(3, 0)$.

4. Maximum and minimum values: The function has a minimum at $x = \frac{1}{2}$. The value of the function at this point is: $f\left( \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 - \frac{1}{2} - 6 = \frac{1}{4} - \frac{1}{2} - 6 = -6.25$ So, the minimum value is $f\left( \frac{1}{2} \right) = -6.25$. The function has no maximum as it increases indefinitely.

Let me know if you'd like more details on any of these steps!

Here are 5 related questions:

1. What is the significance of the slope in Cowling’s rule in terms of dosage calculation?
2. How would the dosage change if the adult dose was 600 mg for the same 4-year-old child?
3. What happens to the child’s dosage as the child’s age increases to 24 years?
4. How do you determine if a function is concave up or concave down using the second derivative?
5. What are the conditions for a quadratic function to have a maximum or minimum?

Tip: For quadratic functions like $f(x) = ax^2 + bx + c$, if $a > 0$, the graph is concave up (minimum), and if $a < 0$, it's concave down (maximum).