Gravity Function E18C6E

1. **State the problem:** We are given the gravitational attraction function $$F(x) = \frac{350}{x^2}$$ where $F$ is in newtons and $x$ is the distance in millions of meters from the moon's surface. 2. **Part A: Graph the function $F(x)$ for $x$ between 0 and 10.** The function is a rational function with a vertical asymptote at $x=0$ and decreases as $x$ increases. 3. **Part B: Describe the behavior of $F$ as $x$ increases.** As $x$ increases, since $F(x) = \frac{350}{x^2}$, the denominator grows larger, so $F(x)$ decreases rapidly towards zero. This means the gravitational attraction weakens as the distance increases. 4. **Part C: Find the average rate of change of $F(x)$ from $x=1$ to $x=h$, where $h>1$.** The average rate of change formula is: $$\text{Average rate of change} = \frac{F(h) - F(1)}{h - 1}$$ Calculate $F(1)$ and $F(h)$: $$F(1) = \frac{350}{1^2} = 350$$ $$F(h) = \frac{350}{h^2}$$ Substitute into the formula: $$\frac{\frac{350}{h^2} - 350}{h - 1} = \frac{350\left(\frac{1}{h^2} - 1\right)}{h - 1} = \frac{350\left(\frac{1 - h^2}{h^2}\right)}{h - 1} = \frac{350(1 - h^2)}{h^2(h - 1)}$$ Factor numerator $1 - h^2$ as $(1 - h)(1 + h)$: $$\frac{350(1 - h)(1 + h)}{h^2(h - 1)}$$ Note that $1 - h = -(h - 1)$, so: $$\frac{350 \cdot -(h - 1)(1 + h)}{h^2(h - 1)} = \frac{-350 (h - 1)(1 + h)}{h^2 (h - 1)}$$ Cancel $(h - 1)$: $$\frac{-350 \cancel{(h - 1)} (1 + h)}{h^2 \cancel{(h - 1)}} = \frac{-350 (1 + h)}{h^2}$$ 5. **Final answer:** The average rate of change of $F(x)$ from $x=1$ to $x=h$ is: $$\boxed{\frac{-350(1 + h)}{h^2}}$$ This shows the rate at which the gravitational force changes with distance between $1$ and $h$ million meters.