Gravity Function 6A74Ce

1. **State the problem:** We are given the gravitational attraction function $$F(x) = \frac{350}{x^2}$$ where $x$ is the distance in millions of meters from the moon's surface, and $F(x)$ is the force in newtons. 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. - The graph is a hyperbola-like curve starting very high near $x=0$ and approaching zero as $x$ approaches 10. 3. **Part B: Describe the behavior of $F$ as $x$ increases.** - Since $F(x) = \frac{350}{x^2}$, as $x$ increases, $x^2$ increases, so $F(x)$ decreases. - This means the gravitational attraction decreases rapidly as the distance from the moon 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: $$\text{Average rate of change} = \frac{F(h) - F(1)}{h - 1}$$ - Substitute $F(x) = \frac{350}{x^2}$: $$= \frac{\frac{350}{h^2} - \frac{350}{1^2}}{h - 1} = \frac{\frac{350}{h^2} - 350}{h - 1}$$ - Find common denominator in numerator: $$= \frac{\frac{350 - 350h^2}{h^2}}{h - 1} = \frac{350(1 - h^2)}{h^2 (h - 1)}$$ - Factor numerator $1 - h^2 = (1 - h)(1 + h)$: $$= \frac{350 (1 - h)(1 + h)}{h^2 (h - 1)}$$ - Notice $1 - h = -(h - 1)$, so: $$= \frac{350 \cdot (-(h - 1)) (1 + h)}{h^2 (h - 1)} = \frac{-350 (h - 1)(1 + h)}{h^2 (h - 1)}$$ - Cancel $(h - 1)$ terms: $$= \frac{-350 \cancel{(h - 1)} (1 + h)}{h^2 \cancel{(h - 1)}} = \frac{-350 (1 + h)}{h^2}$$ - Final simplified average rate of change: $$\boxed{\frac{-350 (1 + h)}{h^2}}$$ This expression shows the average rate of change of the gravitational force between $x=1$ and $x=h$. **Summary:** - The graph is a decreasing curve from very high near zero to near zero at $x=10$. - The gravitational force decreases as distance increases. - The average rate of change from $1$ to $h$ is $$\frac{-350 (1 + h)}{h^2}$$, fully simplified with no nested fractions.